CHAPTER

06
Binomial Theorem
Learning Part
Session 1
● Binomial Theorem for Positive Integral Index

● Pascal’s Triangle

Session 2
● General Term

● Middle Terms

● Greatest Term

● Trinomial Expansion

Session 3
● Two Important Theorems

● Divisibility Problems

Session 4
● Use of Complex Numbers in Binomial Theorem

● Multinomial Theorem

● Use of Differentiation

● Use of Integration

● When Each Term is Summation Contains the Product of Two Binomial Coefficients or Square of Binomial

Coefficients
● Binomial Inside Binomial

● Sum of the Series

Practice Part
● JEE Type Examples
● Chapter Exercises

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Session 1
Binomial Theorem for Positive Integral Index,
Pascal’s Triangle
n
An algebraic expression consisting of two dissimilar terms Cr x n - r ar
2. Let S = ( x + a) n = å
n
with positive or negative sign between them is called a r =0
binomial expressions. Replacing r by n - r , we have
n n
a p q
For example, x + a, x 2 a - , - ,5 - x, S = ( x + a) n = å
n
Cn - r x n - ( n - r ) an - r = å
n
Cn - r x r an - r
2
x x x4 r =0 r =0

1 = n Cn an + n Cn - 1 an - 1 x + n Cn - 2 an - 2 x + ... + n C0 x n
2
( x 2 + 1) 1 / 3 - , etc., are called binomial
( x 3 + 1) Thus, replacing r by n - r , we are infact writing the binomial
expansion in reverse order.
expressions.

Remarks Some Important Points

1. An algebraic expression consisting of three dissimilar terms 1. Replacing a by (- a) in Eq. (i), we get
3
is called a trinomial. e.g. a + 2b + c, x - 2 y + 3z,2a - + g ,
b (x - a )n = nC 0 x n - 0 a 0 - nC1 x n - 1 a 1
etc. are called the trinomials.
+ n C 2 x n - 2 a 2 - ... +...+ ( - 1) r n C r x n - r a r
2. In general, expressions consisting more than two dissimilar
terms are known as multinomial expressions. + ... + ( - 1) n n C n x 0 a n …(ii)
n
or ( x - a ) n = å ( - 1) r n
Cr x n - r a r
Binomial Theorem for r =0

Positive Integral Index 2. On adding Eqs. (i) and (ii), we get

(x + a )n + (x - a )n = 2 { n C 0 x n - 0 a 0
If x , a Î C and n Î N , then
+ n C 2 x n - 2 a 2 + n C 4 x n - 4 a 4 + ... }
( x + a ) n = n C 0 x n - 0 a 0 + n C 1 x n -1 a 1 + n C 2 x n - 2 a 2 + ...
= 2 {Sum of terms at odd places}
+ nCr x n - r a r
The last term is n C n a n or n C n - 1 x a n - 1 ,
+ ... + n C n - 1 x 1 a n - 1 + n C n x 0 a n …(i)
according as n is even or odd, respectively.
n
or ( x + a) = n
å n
Cr x n -r
a r 3. On subtracting Eq. (ii) from Eq. (i), we get
r =0 (x + a )n - (x - a )n = 2 { n C1 x n - 1 a 1
n n n n
Hence, C 0 , C 1 , C 2 ,..., C n are called binomial coefficients. + n C 3 x n - 3 a 3 + n C 5 x n - 5 a 5 + ... }

Remark = 2 {Sum of terms at even places}

1. In each term, the degree is n and the coefficient of x n - r ar is The last term is n C n - 1 x a n - 1 or n C n a n ,
equal to the number of ways x, x, x, ... , x , a, a, a, ... , a according as n is even or odd, respectively.
14243 14243
can be arranged, which ( n - r ) times r times
n! 4. Replacing x by 1 and a by x in Eq. (i), we get
is given by = n Cr
( n - r )! r ! (1 + x ) n = n C 0 x 0 + n C 1 x 1 + n C 2 x 2
5! 5 0 5! 4 5! 3 2
For example, ( x + a)5 = x a + x a+ x a + ... + n C r x r + ... + n C n - 1 x n -1
5! 0 ! 4 ! 1! 3! 2!
5! 2 3 5! 5! 0 5 + n C n x n ...(iii)
+ x a + x a4 + x a n
2! 3! 1! 4 ! 0 ! 5!
5 5 5 4 5 3 2
= C0 x + C1 x a + C2 x a + C3 x a + C4 x a + C5 a 5 2 3 5 4 5 5
or (1 + x ) n = å n Cr xr
r =0
 Chap 06 Binomial Theorem 439

5. Replacing x by (- x) in Eq. (iii), we get = P + Q (given) …(i)

and ( x - a )n = n C 0 x n - 0 a 0 - n C 1 x n - 1 a1 +n C 2 x n - 2 a 2
(1 - x ) n = n C 0 x 0 - n C 1 x 1 + n C 2 x 2
- n C 3 x n - 3 a 3 + ... + n C n x n - n an
- ... + ( - 1) r n C r x r + ... + n C n ( - 1) n x n
= (n C 0 x n + n C 2 x n - 2 a 2 + n C 4 x n - 4 a 4 + ...)
n
or (1 - x ) n = å ( - 1) r n
Cr x r - (n C 1 x n - 1 a + n C 3 x n - 3 a 3 +n C 5 x n - 5 a 5 + ...)
r =0
= P - Q (given) ...(ii)
5
y Example 1. Expand æç 2a - ö÷ by binomial theorem.
3 (i) P 2 - Q 2 = ( P + Q ) ( P - Q )
è bø
= ( x + a )n × ( x - a )n
Sol. Using binomial theorem, we get
5 0 1 = ( x 2 - a 2 )n [from Eqs. (i) and (ii)]
æ 3ö 5 5-0 æ 3ö 5 5 -1 æ 3ö
ç2a - ÷ = C 0 (2a ) ç - ÷ + C 1(2a ) ç- ÷
è bø è bø è bø (ii) ( x + a )2n - ( x - a )2n = [( x + a )n ]2 - [( x - a )n ]2
2 3
æ 3ö æ 3ö = ( P + Q )2 - ( P - Q )2
+ 5C 2 (2a )5 - 2 ç - ÷ + 5C 3 (2a )5 - 3 ç- ÷
è bø è bø = 4PQ [from Eqs. (i) and (ii)]
4 5
5 5-4 æ 3ö 5 5 - 5 æ 3ö
+ C 4 (2a ) ç - ÷ + C 5 (2a ) ç- ÷
è bø è bø y Example 4. Show that (101) 50 > (100) 50 + (99 ) 50 .
2
æ3ö æ3ö Sol. Since, (101)50 - (99 )50 = (100 + 1)50 - (100 - 1)50
= 5C 0 (2a )5 - 5C 1 (2a )4 ç ÷ + 5C 2 (2a )3 ç ÷
èb ø èb ø
3 4 5
= 2 { 50 C 1 (100)49 + 50
C 3 (100)47 + 50
C 5 (100)45 + ...}
æ3ö æ3ö æ3ö
- 5C 3 (2a )2 ç ÷ + 5C 4 (2a )1 5
ç ÷ - C5 ç ÷ =2´ 50
C 1 (100)49 + 2 { 50C 3 (100)47 + 50
C 5 (100)45 + ...}
èb ø èb ø èb ø
240 a 4 720 a 3 1080 a 2 810 a 243 = (100)50 + ( a positive number) > (100)50
= 32a 5 - + - + 4 - 5
b b2 b3 b b Hence, (101)50 - (99 )50 > (100)50

y Example 2. Simplify Þ (101)50 > (100)50 + (99 )50

( x + ( x 2 - 1 )) 6 + ( x - ( x 2 - 1 ) 6 . n
1
2
y Example 5. If an = å n
Cr
, find the
Sol. Let (x - 1) = a r =0
n
Then, ( x + a )6 + ( x - a )6 = 2 { 6 C 0 x 6 - 0 a 0 + 6C 2 x 6 - 2 a 2 r
value of å n
Cr
×
+ 6C 4 x 6 - 4 a 4 + 6C 6 x 6 - 6 a 6 } r =0
n
= 2 { x 6 + 15x 4a 2 + 15x 2a 4 + a 6 } r
[from point (2)] Sol. Let P = å n
Cr
…(i)
6 4 2 2 2 2 2 3 r =0
= 2 { x + 15x ( x - 1) + 15x ( x - 1) + ( x - 1) }
[Qa = x 2 - 1 ] Replacing r by (n - r ) in Eq. (i), we get
n
(n - r ) n
(n - r )
= 2(32x 6 - 48x 4 + 18x 2
- 1) P= å n
Cn - r
= å n
Cr
[Q n C r = n C n - r ] …(ii)
r =0 r =0

y Example 3. In the expansion of ( x + a )n , if sum of On adding Eqs. (i) and (ii), we get
odd terms is P and sum of even terms is Q, prove that n
n n
1
(i) P 2 - Q 2 = ( x 2 - a 2 )n
2P = å n
Cr
=n å n
Cr
= nan [given]
r =0 r =0
(ii) 4 PQ = ( x + a ) 2n - ( x - a ) 2n n
\ P= an
Sol. Q ( x + a )n = n C 0 x n - 0a 0 + n C 1 x n - 1 a1 + n C 2 x n - 2 a 2 2
n
+ n C 3 x n - 3 a 3 + ... + ... + n C n x n - n an r n
Hence, å n
Cr
=
2
an
= (n C 0 x n + n C 2 x n - 2 a 2 +n C 4 x n - 4 a 4 + ...) r =0

+ (n C 1 x n - 1 a1 + n C 3 x n - 3 a 3 + n C 5 x n - 5 a 5 + ...)
440 Textbook of Algebra

Properties of Binomial How to Construct a Pascal’s Triangle

Expansion ( x + a )n Binomial coefficients in the expansion of ( x + a ) 3 are
(i) This expansion has (n + 1) terms. 1 3 3 1

1 3 3 1
(ii) Since, nC r = nC n - r , we have
1 (1 + 3 ) (3 + 3 ) (3 + 1) 1
n
C 0 = n Cn = 1 Then, 1 4 6 4 1
n n
C1 = Cn - 1 = n are the binomial coefficients in the expansion of ( x + a ) 4 .
n n (n - 1)
C2 = n Cn - 2 =
and so on. y Example 6. Find the number of dissimilar terms
2!
in the expansion of (1 - 3x + 3x 2 - x 3 ) 33 .
(iii) In any term, the suffix of C is equal to the index of a
and the index of x = n - (suffix of C ). Sol. (1 - 3x + 3x 2 - x 3 )33 = [(1 - x )3 ]33 = (1 - x )99
(iv) In each term, sum of the indices of x and a is equal to n. Therefore, number of dissimilar terms in the expansion of
(1 - 3x + 3x 2 - x 3 )3 is 100.
Properties of Binomial Coefficient n
r × nC r
n ænö
(i) C r can also be represented by C (n, r ) or ç ÷ . y Example 7. Find the value of å nC .
èr ø r =1 r -1
n
(ii) n C x = nC y , then either x = y or n = x + y . Cr n -r +1
Sol. Q n
=
Cr -1 r
n!
So, n C r = n C n - r = n
r ! (n - r ) ! r × Cr
\ n
= ( n - r + 1)
n +1 Cr
(iii) n C r + nC r - 1 = Cr -1
n
r × nC r n n n
å nC å ( n - r + 1) = å ( n + 1) - å r
n
Cr n -r +1 \ =
(iv) = r =1 r -1 r =1 r =1 r =1
n r
C r -1
n

(v) n C r =
n
× n -1
Cr - 1
= ( n + 1) å1 - (1 + 2 + 3 + ... + n )
r =1
r
n ( n + 1) n ( n + 1)
= ( n + 1) × n - =
2 2
Pascal’s Triangle y Example 8. Let C r stands for n C r , prove that
Coefficients of binomial expansion can also be easily
determined by Pascal’s triangle. (C 0 + C 1 ) (C 1 + C 2 ) (C 2 + C 3 ) ...(C n -1 + C n )

(x + a)0 1 (n + 1)n
= (C 0C 1C 2 ... C n -1 ).
(x + a)1 1 1 n!
(x + a)2 1 2 1 Sol. LHS = (C 0 + C 1 ) (C 1 + C 2 ) (C 2 + C 3 ) ... (C n - 1 + C n )
n n
(x + a)3 3 3
Õ(C r - 1 + C r ) = Õ(n + 1C r )
1 1 n +1
= [Q n C r + n C r -1 = Cr ]
(x + a)4 1 4 6 4 1 r =1 r =1

( x + a )5 n
æn + 1ö n é n n n -1 ù
1 5 10 10 5 1
= Õ çè r ø
÷ Cr -1 êëQ C r = r × Cr - 1ú
û
Pascal triangle gives the direct binomial coefficients. r =1
n n n
1
For example, = Õ ( n + 1) × Õ r × Õ C r - 1
( x + a ) 4 = 1 × x 4 + 4 × x 3 × a + 6 × x 2a 2 r =1 r =1 r =1

3 4 1
+ 4 × x a + 1× a n
= (n + 1) × × (C 0C 1C 2 ... C n - 1 )
n!
= x 4 + 4 x 3a + 6 x 2a 2 + 4 x a 3 + a 4 (n + 1)n
= (C 0 C 1 C 2 ... C n - 1 ) = RHS
n!
 Chap 06 Binomial Theorem 441

y Example 9. Find the sum of the series æ 1ö

n
æ 3ö æ
n
7ö
n
æ 15 ö
n
= ç1 - ÷ + ç1 - ÷ + ç1 - ÷ + ç1 - ÷
n ì1 3r 7r 15r ü è 2 ø è 4 ø è 8 ø è 16 ø
å ( - 1)r nC r í 2r + 2r
+ 3r
+ 4r
+ ... upto m termsý.
r =0 î 2 2 2 þ + ... upto m terms
n
å ( - 1)
n r n r n 2n 3n 4n
Sol. Q(1 - x ) = Cr x ...(i) æ1ö æ1ö æ1ö æ1ö
r =0
=ç ÷ +ç ÷ +ç ÷ +ç ÷ + ... upto m terms
n ìï æ 1 ö r r r è2ø è2ø è2ø è2ø
æ3ö æ7 ö
Let P= å ( - 1) r n
Cr íç ÷ + ç ÷ + ç ÷
è ø è ø è8ø n é ìï æ 1 ön üï ù
m
2 4
r =0 îï æ1ö ê
üï ç ÷ 1 - íç ÷ ý ú
r
æ 15 ö è2ø ê è2ø ï ú
+ ç ÷ + ... upto m termsý
è 16 ø ïþ ë îï þ û
n r n r =
æ1ö æ3ö æ1ö
n
= å ( - 1)r n
Cr × ç ÷
è2ø
+ å ( - 1) C r × ç ÷
r n
è4ø 1- ç ÷
r =0 r =0
n rn r è2ø
Cr × æç ö÷ + å ( - 1)r nCr × æç ö÷
7 15
å (- 1)
r n
+
è8ø r = 0 è 16 ø (2mn - 1)
r =0 =
+ K upto m terms 2mn (2n - 1)

#L Exercise for Session 1

10
1. The value of å r × 10Cr × 3r × ( -2)10 - r is
r =0

(a) 10 (b) 20 (c) 30 (d) 300

15
æ 1 1ö
2. The number of dissimilar terms in the expansion of ç x + + x 2 + 2 ÷
è
are
x x ø
(a) 61 (b) 121 (c) 255 (d) 16

3. 3 1/ 2 5
The expansion {x + ( x - 1) } + {x - ( x - 1) } is a polynomial of degree 3 1/ 2 5

(a) 5 (b) 6 (c) 7 (d) 8

4. 6
( 2 + 1) - ( 2 - 1) is equal to 6

(a) 101 (b) 70 2 (c) 140 2 (d) 120 2

5. The total number of dissimilar terms in the expansion of ( x + a ) 100

+ (x - a ) 100
after simplification will be
(a) 202 (b) 51
(c) 50 (d) 101

6. The number of non-zero terms in the expansion of (1+ 3 2x )9 + (1- 3 2x )9, is

(a) 0 (b) 5
(c) 9 (d) 10
n
æ C1 ö æ C2 ö æ Cn ö
7. If (1 + x )n = å Cr x r , ç1 +
è
÷ ç1 +
C0 ø è
÷ ... ç1 +
C1 ø è
÷ is equal to
Cn - 1 ø
r =0

n n -1
(n + 1)n - 1
(a) (b)
(n - 1)! (n - 1) !
(n + 1)n (n + 1)n + 1
(c) (d)
n! n!
n+1 n -1
8. If Cr + 1 : nCr : Cr - 1 = 11: 6 : 3, nr is equal to
(a) 20 (b) 30
(c) 0 (d) 50
Session 2
General Term, Middle Terms, Greatest Term,
Trinomial Expansion

General Term (ii) the coefficient of x - 7 in the expansion of

11
The term n C r x n - r a r is the (r + 1) th term from æ 1 ö
ç ax - 2 ÷ .
è bx ø
beginning in the expansion of ( x + a ) n . It is usually called
the general term and it is denoted by Tr + 1 . Also, find the relation between a and b, so that these
i.e., Tr + 1 = nC r x n - r a r coefficients are equal.
11
æ 1 ö
y Example 10. Find the 7th term in the expansion of Sol. (i) Here, Tr +1 = 11C r (ax 2 )11- r ç ÷
è bx ø
13
æ 1 ö
ç4 x - ÷ . a11 - r
= 11C r × × x 22 - 3r …(i)
è 2 xø br

6
æ 1 ö Now, in order to find out the coefficient of x 7 , 22 - 3r
Sol. Seventh term, T 7 = T 6 + 1 = 13C 6 ( 4 x )13 - 6 ç - ÷
è 2 xø must be 7,
1 i.e. 22 - 3r = 7
= 13C 6 × 4 7 × x 7 × 6 3
2 ×x \ r =5
= 13C 6 × 28 × x 4 Hence, putting r = 5 in Eq. (i), we get
a6
8 Required coefficient = 11C 5 .
y Example 11. Find the coefficient of x in the b5
10
æ 1ö æ 1 ö
R
expansion of ç x 2 - ÷ . (ii) Here, T R + 1 = 11C R (ax )11 - R ç - 2 ÷
è xø è bx ø
r
æ 1ö R
Sol. Here, Tr = 10C r ( x 2 )10 - r ç - ÷ æ 1ö
+1
è xø = C R (a )11 - R ç - ÷ × x 11 - 3 R
11
è bø
1
= 10C r x 20 - 2r × ( - 1)r × a11 - R
xr = ( - 1)R × 11C R × × x 11 - 3 R …(ii)
10 r 20 - 3r bR
= C r ( - 1) × x ...(i)
Now, in order to find out the coefficient of x - 7 ,
8
Now, in order to find out the coefficient of x , 20 - 3r must 11 - 3R must be - 7.
be 8. i.e., 11 - 3R = - 7 Þ R = 6. Hence, putting R = 6 in Eq.
i.e. 20 - 3r = 8 (ii), we get
\ r =4 Required coefficient
Hence, putting r = 4 in Eq. (i), we get a5 a5
= ( - 1)6 × 11C 6 × = 11C 5 × [Q n C r = n C n - r ]
10 × 9 × 8 × 7 b 6
b 6
Required coefficient = ( - 1)4 × 10C 4 = = 210
1×2×3× 4 Also given, coefficient of x 7 in
11 11
æ 2 1 ö -7 æ 1 ö
y Example 12. Find ç ax + ÷ = coefficient of x in ç ax - ÷
è bx ø è bx 2 ø
(i) the coefficient of x 7 in the expansion of
11 11 a6 a5
æ 2 1ö Þ C5 × = 11C 5 ×
Þ ab = 1
ç ax + ÷ . b5 b6
è bx ø which is the required relation between a and b.
 Chap 06 Binomial Theorem 443

y Example 13. Find the term independent of x in the

9
How to Find Free from Radical Terms or
expansion of æç x 2 - ö÷ .
3 1 Rational Terms in the Expansion of
è2 3x ø ( a1/ p + b1/q )N , " a, b Î Prime Numbers
9 -r r
æ3 ö æ 1 ö -r
Sol. Here, Tr = 9C r ç x 2 ÷ ç- ÷
First, find Tr + 1 = N
C r (a 1 /p ) N (b 1 /q ) r
+1
è2 ø è 3x ø - r ) /p
9 -r r
\ Tr + 1 = N
Cr × a ( N × b r /q
æ3ö æ1ö
= ( - 1)r × 9C r × ç ÷ × ç ÷ × x 18 - 3r …(i) By inspection, putting the values of 0 £ r £ N , when
è2ø è3ø
indices of a and b are integers.
If this term is independent of x, then the index of x must be
zero, i.e., 18 - 3r = 0 Þ r = 6 Remark
Therefore, (r + 1) th term, i.e., 7th term is independent of x 1. If indices of a and b are positive integers.
and its value by putting r = 6 in Eq. (i) Then, free from radical terms = Terms which are integers
3 6
æ3ö æ1ö 1 \ Number of non-integral terms = Total terms - Number of
= ( - 1)6 × 9C 6 × ç ÷ × ç ÷ = 9C 3 × 3 3 integral terms
è2ø è3ø 2 ×3
2. If indices of a and b both are not positive integers.
9 ×8×7 7 Then, free from radical terms = Rational terms - Integral
= =
( 1 × 2 × 3) 23 × 33 18 terms
3. Number of irrational terms = Total terms – Number of
rational terms
( p + 1) th Term From End in the
Expansion of ( x + a )n y Example 16. Find the number of terms in the
expansion of ( 4 9 + 6 8 ) 500 which are integers.
( p + 1) th term from end in the expansion of ( x + a ) n
n
Sol. Since, ( 4 9 + 6 8 )500 = (91/ 4 + 81/ 6 )500 = (31/ 2 + 21/ 2 )500
= ( p + 1) th term from beginning in the expansion of (a + x )
[Qa, b Îprime numbers]
= nCp a n - p x p
\ General term, Tr +1 = 500
C r (31/ 2 )500 - r × (21/ 2 )r
500 - r
y Example 14. Find the 4th term from the end in the 500
7 = Cr × 3 2 × 2r / 2
æx3 2 ö
expansion of ç - 2÷ . = 500
C r × 3250 - r / 2 × 2 r / 2
è 2 x ø 7 Now, 0 £ r £ 500
æx3 2 ö
Sol. 4th term from the end in the expansion of ç - 2÷ For r = 0 , 2, 4, 6, 8, ... , 500, indices of 3 and 2 are positive
è 2 x ø
integers.
= 4th term from beginning in the expansion of
æ 2
7 Hence, number of terms which are integers = 250 + 1 = 251
x3 ö
ç- 2 + ÷
è x 2 ø
y Example 17. Find the sum of all rational terms in
7-3 3
æ 2 ö æx3 ö 7 × 6 × 5 24 x 9 the expansion of ( 31/ 5 + 21/ 3 )15 .
= 7C 3 ç - 2 ÷ ç ÷ = × × = 70x
è x ø è 2 ø 1 × 2 × 3 x 8 23
Sol. The general term in the expansion of (31/ 5 + 21/ 3 )15 is
Tr +1 = 15C r (31/ 5 )15 - r × (21 / 3 )r
y Example 15. Find the (n + 1)th term from the end in
r r
3n 3-
æ 1ö = 15C r × 3 5 ×23
the expansion of ç 2x - ÷ .
è xø Now, 0 £ r £ 15
3n
æ 1ö Forr = 0, 15
Sol. (n + 1)th term from the end in the expansion of ç2x - ÷
è xø Rational terms are T 0 + 1 and T15 + 1.
= (n + 1) th 3term
n
from beginning in the expansion of Then, T 0 + 1 = 15C 0 × 33 × 20 = 27
æ 1 ö
ç - + 2x ÷ and T15 + 1 = 15C 15 × 30 × 25 = 32
è x ø
3n - n
æ 1ö \ Sum of all rational terms = 27 + 32 = 59
= Tn + 1 = 3n C n ç - ÷ (2x )n = 3n C n × 2n × x - n
è xø
444 Textbook of Algebra

y Example 18. Find the number of irrational terms in y Example 20. If a, b , c and d are any four
the expansion of ( 8 5 + 6 2 )100 . consecutive coefficients in the expansion of (1 + x )n ,
Sol. Since, ( 8 5 + 6 2 )100 = (51/ 8 + 21/ 6 )100 then prove that:
\ General term, Tr = 100
C r (51 / 8 )100 - r (21 / 6 )r a c 2b
+1 (i) + = .
= 100
C r (5)(100 - r ) / 8 × (2)r / 6
a+b c + d b + c
2
As, 2 and 5 are coprime. æ b ö ac
(ii) ç ÷ > , if x > 0.
\ Tr + 1 will be rational, if (100 - r ) is a multiple of 8 and r is èb + c ø (a + b ) (c + d )
a multiple of 6.
Sol. Let a, b, c and d be the coefficients of the r th, (r + 1)th,
Also, 0 £ r £ 100
(r + 2)th and (r + 3)th terms respectively, in the expansion
\ r = 0, 6, 12, 18, ..., 96
of (1 + x )n . Then,
Now, 100 - r = 4, 10, 16, ... , 100 …(i) -1
Tr = Tr -1+1 = nC r -1 xr
and 100 - r = 0, 8, 16, 24, ... , 100 …(ii)
The common terms in Eqs. (i) and (ii) are 16, 40, 64 and 88. \ a = nC r -1 …(i)
\ r = 84, 60, 36, 12 gives rational terms. Q Tr = nC r x r
+1
\ The number of irrational terms = 101 - 4 = 97
\ b = nC r …(ii)
n r +1
Problems Regarding Three/Four Q Tr +2 = T (r + 1) + 1 = Cr +1 x

Consecutive Terms or Coefficients \ c = nC r +1 ….(iii)

n r +2
and Tr = T (r = Cr x
(i) If consecutive coefficients are given +3 + 2) + 1 +2

In this case, divide consecutive coefficients pairwise, we \ d = nC r +2 …(iv)

get equations and then solve them. From Eqs. (i) and (ii), we get
a + b = n C r - 1 + n C r = n + 1C r
y Example 19. Let n be a positive integer. If the n +1 n æn + 1ö
coefficients of rth, (r + 1)th and (r + 2)th terms in the = × Cr - 1 = ç ÷a
r è r ø
expansion of (1 + x )n are in AP, then find the a r
relation between n and r. \ = …(v)
a+b n +1
n r -1
Sol. Q Tr = T (r - 1) + 1 = Cr -1 x From Eqs. (ii) and (iii), we get
n r n r +1 n +1
Tr +1 = Cr x and Tr +2 = T (r + 1) +1 = Cr +1 x b + c = nC r + nC r +1 = Cr +1

\ Coefficients of rth, (r + 1)th and (r + 2)th terms in the æn + 1ö n æn + 1ö

expansion of =ç ÷ Cr = ç ÷b
èr + 1ø èr + 1ø
(1 + x )n are n C r - 1, n C r , n C r + 1.
b r +1
Q Given, n C r - 1,
n
Cr , n Cr +1 are in AP. \ = …(vi)
b+c n +1
and n ³r +1 From Eqs. (iii) and (iv), we get
n n
Cr Cr n +1
\
-1
, 1,
+1
are also in AP. c + d = nC r + 1 + nC r +2 = Cr +2
n n
Cr Cr æn + 1ö n æn + 1ö
r n -r =ç ÷ Cr +1 =ç ÷c
Þ , 1, are in AP. èr + 2ø èr + 2ø
n -r +1 r +1 c r +2
r n -r n - 2r +1 n - 2r - 1 \ = …(vii)
Þ 1- = -1 Þ = c +d n +1
n -r +1 r +1 n -r +1 r +1
From Eqs. (v), (vi) and (vii), we get
Þ nr - 2r 2 + r + n - 2r + 1 a b c
, and are in AP.
= n 2 - 2nr - n - nr + 2r 2 + r + n - 2r - 1 a+b b+c c +d
Þ n 2 - 4nr + 4r 2 = n + 2 Þ (n - 2r )2 = n + 2 a c æ b ö
(i) + =2ç ÷
Corollary I For r = 2, n = 7 [Qn ³ 3] a+b c +d èb + c ø

Corollary II For r = 5, n = 7, 14 [Qn ³ 6] a c 2b

or + =
a+b c +d b+c
 Chap 06 Binomial Theorem 445

(ii) AM > GM From Eqs. (i) and (vi), we get

æ b ö æ a öæ c ö 5 3
\ C 1 × x 4 × y = 240 Þ 5× x4 × x = 240
ç ÷> ç ÷ç ÷ 2
èb + c ø èa + b ø èc + d ø
5
2 \ x = 32 = 25 Þ x = 2
æ b ö ac
Þ ç ÷ > From Eq. (vi), we get y = 3
èb + c ø (a + b ) (c + d )
Hence, x = 2, y = 3 and n = 5
Remembering Method
Q
a b c d
Middle Terms
a+b b+c The middle term depends upon the value of n.
c+d
(i) When n is even The total number of terms in the
a b c expansion of ( x + a ) n is n + 1 (odd). So, there is only
\ , and are in AP.
a+b b+c c +d æn ö
one middle term, i.e., ç + 1÷ th term is the middle
è2 ø
(ii) If consecutive terms are given
term. It is given byTn /2 + 1 = n C n /2 x n /2 a n /2
In this case, divide consecutive terms pairwise. i.e., If four
(ii) When n is odd The total number of terms in the
consecutive terms are Tr , Tr + 1 , Tr + 2 , Tr + 3 . Then, find expansion of ( x + a ) n is n + 1 (even). So, there are
Tr + 1 Tr + 2 Tr + 3
, , Þ l1 , l2 , l 3 (say). Then, divide l2 by æ n + 1ö æn + 3ö
Tr Tr + 1 Tr + 2 two middle terms, i.e., ç ÷ th and ç ÷ th are
è 2 ø è 2 ø
l1 and l 3 by l2 and solve. two middle terms. They are given by
n +1 n -1
y Example 21. If the 2nd, 3rd and 4th terms in the T n + 1 =T æ n - 1 ö = nC n - 1 × x 2 ×a 2
ç ÷ +1
expansion of ( x + y )n are 240, 720 and 1080 2 è 2 ø 2
n -1 n +1
respectively, find x , y and n. n
and T n + 3 = T æ n + 1 ö = Cn +1 × x 2 ×a 2
Sol. Given, T 2 = T1 + 1 = n C 1 × x n - 1× y = 240 …(i) ç ÷ +1
2 è 2 ø 2
T 3 = T 2 + 1 = n C 2 × x n - 2 × y 2 = 720 …(ii)
and T 4 = T 3 + 1 = n C 3 × x n - 3 × y 3 = 1080 …(iii) y Example 22. Find the middle term in the
12
On dividing Eq. (ii) by Eq. (i), we get æa ö
expansion of ç + bx ÷ .
n
C 2 × x n - 2 × y 2 720 èx ø
=
C1 × x n - 1 × y
n 12
240 æa ö
Sol. The number of terms in the expansion of ç + bx ÷ is 13
æn - 2 + 1ö y y 6 èx ø
Þ ç ÷× =3 Þ = …(iv) æ 12 ö
è 2 ø x x n -1 (odd), its middle term is ç + 1÷ th, i.e., 7th term.
è2 ø
Also, dividing Eq. (iii) by Eq. (ii), we get 6
12 æa ö
n
C3 × x n-3
×y 3
1080 \ Required term, T 7 = T 6 +1 = C 6 ç ÷ (bx )6
= èx ø
n-2
n
C2 × x ×y 2
720 = 12 C 6 a 6 b 6 = 924 a 6 b 6
æn - 3 + 1ö y 3 y 9
Þ ç ÷× = Þ = …(v) y Example 23. Find the middle term in the expansion
è 3 ø x 2 x 2 ( n - 2) 9
æ x3 ö
From Eqs. (iv) and (v), we get of ç 3x - ÷ .
6 9 è 6 ø
= æ
9
x3 ö
n - 1 2 ( n - 2) Sol. The number of terms in the expansion of ç3x - ÷ is
10 (even). So, there are two middle terms, è
6 ø
Þ 12n - 24 = 9 n - 9
Þ 3n = 15 æ9 +1ö æ9 + 3ö
i.e. ç ÷ th and ç ÷ th terms. They are given by T 5
\ n =5 è 2 ø è 2 ø
3 and T 6 .
From Eq. (iv), we get y = x …(vi)
2
446 Textbook of Algebra

4
æ x 3ö Now, on substituting values of n, x and a in Eq. (i), we get
\ T 5 = T 4 +1 = 9C 4 (3x )5 ç - ÷
è 6 ø r £ m + f or r £ m
where, m Î N and 0 < f < 1
9 × 8 × 7 × 6 5 5 x 12 189 17
= ×3 x × 4 = x In the first case,Tm +1 is the greatest term, while in the
1× 2× 3× 4 6 8
second case, Tm and Tm + 1 are the greatest terms and both
5
æ x 3ö are equal (numerically).
and T 6 = T 5 + 1 = 9C 5 (3x )4 ç - ÷
è 6 ø Shortcut Method
x 15 To find the greatest term (numerically) in the expansion of
= - 9C 4 × 34 × x 4 ×
65 (x + a )n .
n
9 × 8 × 7 × 6 4 x 19 21 19 æ xö
=- ×3 × 5 = - x Now, (x + a )n = a n ç1 + ÷
1× 2× 3× 4 6 16 è aø
x
y Example 24. Show that the middle term in the (n + 1)
a
expansion of (1 + x ) 2n is Calculate m=
æ x ö
1 × 3 × 5 ...(2n - 1) n n ç +1 ÷
× 2 x , n being a positive integer. è a ø
n!
Case I If m ÎInteger, then Tm and Tm + 1 are the greatest terms
Sol. The number of terms in the expansion of (1 + x )2n is
and both are equal (numerically).
2n + 1 (odd), its middle term is (n + 1)th term.
Case II If m Ï Integer, then T[m ] + 1 is the greatest term, where
\ Required term = Tn + 1
[ × ] denotes the greatest integer function.
2n ! n (1 × 2 × 3 × 4 . .. (2n -1) × 2n ) n
= 2n C n x n = x = x
n! n! n! n! y Example 25. Find numerically the greatest term in
{1 × 3 × 5 ... (2n - 1)} {2 × 4 × 6... 2n } n the expansion of (2 + 3x ) 9 , when x = 3 / 2.
= x
n! n! Sol. Let Tr be the greatest term in the expansion of
+1
{1 × 3 × 5... (2n - 1)} 2n (1 × 2 × 3... n ) n (2 + 3x )9 , we have
= x
n! n! Tr +1 æ 9 - r + 1 ö 3x æ 10 - r ö 3 3 90 - 9r
=ç ÷ =ç ÷ ´ =
n
{1 × 3 × 5... (2n - 1)} 2 n ! n 1 × 3 × 5... (2n - 1) n n Tr è r ø 2 è r ø 2 2 4r
= x = 2 x
n! n! n! [Q x = 3 / 2]
Tr + 1
\ ³1
Tr
Greatest Term Þ
90 - 9r
³ 1 Þ 90 ³ 13r
If Tr and Tr +1 are the rth and (r + 1) th terms in the 4r
expansion of ( x + a ) n , then 90 12
\ r £ =6
13 13
Tr + 1 n
C r × x n -r × a r æ n - r + 1ö a
= =ç ÷× or r £6
12
Tr n
Cr - 1 × x n -r +1
×a r -1 è r ø x 13
Let numerically, Tr +1 be the greatest term in the above \ Maximum value of r is 6.
expansion. Then, So, greatest term = T 6 + 1 = 9C 6 (2)9 - 6 (3x )6
Tr + 1 æ n - r + 1ö a æ 3ö
6
Tr + 1 ³ Tr or ³1 Þ ç ÷ ³1 = 9C 3 × 23 × ç3 ´ ÷
Tr è r ø x è 2ø
[Qa may be + ve or - ve] 9 × 8 × 7 23 × 312 7 ´ 313
(n + 1) = × =
or r£ …(i) 1 × 2× 3 26 2
æ x ö 9
ç1 + ÷ æ
Aliter Since, (2 + 3x )9 = 29 ç1 +
3x ö
è a ø ÷
è 2 ø
 Chap 06 Binomial Theorem 447

( 9 + 1)9 3x \ Greatest term (when r = 2) = 11C 2 (3)9 ( - 5x )2

10 ´
Now, m = =
2
4 [Q x = 3 / 2] é 1ù
= 11C 2 (3)9 ( - 1)2 êëQ x = 5 úû
3x 9
+1 +1
2 4 11 × 10 9
= × 3 = 55 ´ 39
90 12 1×2
= =6 ¹ Integer
13 13 and greatest term (when r = 3) = 11C 3 (3)8 ( - 5x )3
\ The greatest term in the expansion is 11 é 1ù
9 = C 3 ( 3) 8 ( - 1) 3 êëQ x = 5 úû
T[m ] + 1 = T 6 + 1 in (2 + 3x )
11 × 10 × 9 8
æ 32 ö
6
= × 3 = 55 ´ 39
9
= C 6 ( 2) 9-6 6 9
( 3x ) = C 3 × 2 × ç ÷3
[Q x = 3 / 2] 1×2×3
è2ø
9 × 8 × 7 312 7 ´ 313
= ×
1 × 2 × 3 23
=
2
Greatest Coefficient
(i) If n is even, then greatest coefficient is n C n /2 ×
y Example 26. Find numerically the greatest term in (ii) If n is odd, then greatest coefficients are n C (n - 1 ) /2 and
1
the expansion of ( 3 - 5x )11 , when x = . n
C (n + 1 ) /2 ×
5
Sol. Let Tr +1 be the greatest term in the expansion of
11
y Example 27. Show that, if the greatest term in the
(3 - 5x ) , we have expansion of (1 + x ) 2n has also the greatest coefficient,
Tr + 1 æ 11 - r + 1 ö 5x n n+1
=ç ÷ -
Tr è r ø 3 then x lies between and .
n+1 n
æ 12 - r ö 1 12 - r
=ç ÷ - = [Q x = 1 / 5] Sol. In the expansion of (1 + x )2n , the middle term is
è r ø 3 3r
æ 2n ö
Tr + 1 12 - r ç + 1÷ th
\ ³1 Þ ³ 1 Þ 12 ³ 4r è2 ø
Tr 3r
i.e., (n + 1) th term, we know that from binomial expan-
\ r £ 3 Þ r = 2, 3
sion, middle term has greatest coefficient.
So, the greatest terms are T 2 + 1 and T 3 + 1. [Q Terms T1, T 2 , T 3 , ..., Tn , Tn + 1, Tn + 2 , ...]
\Greatest term (when r = 2) = T 2 + 1 = 11C 2 (3)9 ( - 5x )2 \ Tn < Tn + 1 > Tn + 2
11 × 10 9 Tn + 1 2n
Cn × x n 2n - n + 1
= × 3 × (1)2 = 55 ´ 39 [Q x = 1 / 5] Þ = = ×x
1×2 Tn 2n
Cn - 1 × x n - 1 n
and greatest term (when r = 3) = T 3 + 1
Tn + 1 n +1
= 11
C 3 ( 3) 8 ( - 5x ) 3 = 11
C 3 ( 3) 8 ( - 1) 3 [Q x = 1 / 5] Þ > 1 or ×x >1
Tn n
11 × 10 × 9 8 n
= 11C 3 × 38 = × 3 = 55 ´ 39 or x> …(i)
1×2×3 n +1
From above, we say that the values of both greatest terms Tn + 2 2n
Cn + 1 x n + 1 2n - (n + 1) + 1
are equal. and = = ×x
Tn + 1 2n
Cn x n
n +1
Aliter 11 n
æ 5x ö = ×x
Since, (3 - 5x )11 = 311 ç1 - ÷ n +1
è 3 ø Tn + 2 n n +1
5x 1 Þ <1 Þ × x < 1 or x < …(ii)
(11 + 1) - 12 ´ - Tn + 1 n +1 n
3 3 é 1ù
Now, m= = êëQ x = 5 úû From Eqs. (i) and (ii), we get
5x 1
- +1 - +1 n n +1
3 3 <x<
n +1 n
4
=3= Corollary For n = 5
1
+1 5 6
3 <x<
Since, the greatest terms in the expansion are T 3 and T 4 . 6 5
448 Textbook of Algebra

Important Properties of Trinomial Expansion

the Binomial Coefficients 2n
n
In the binomial expansion of (1 + x ) . Let us denote the
For n Î N , (1 + x + x 2 ) n = åar x r
r =0
coefficients n C 0 , n C 1 , n C 2 , ..., n C r , ... , n C n by C 0 , C 1 , C 2 , = a 0 + a1 x + a2 x 2
+ K + an x n + K + a 2n x 2n …(i)
... , C r , ... , C n , respectively.
There are (2n + 1) terms. The middle coefficient is a n
(i) The coefficients of the terms equidistant from which is also the greatest.
the beginning and the end are equal
a 0 = a 2n , a 1 = a 2n - 1 , K, a r = a 2n - r
The (r + 1)th term from the beginning in the
expansion of (1 + x ) n is nC r x r × The coefficients of (1 + x + x 2 ) n for n = 0, 1, 2, … can be
arranged in a triangle.
\ The coefficient of the (r + 1) th term from the 1
beginning is nC r and the (r + 1) th term from the end 1 1 1
n 1 2 3 2 1
in the expansion of (1 + x ) = (r + 1) th term from the 1 3 6 7 6 3 1
beginning in the expansion of ( x + 1) n = nC r x n - r 1 4 10 16 19 16 10 4 1
\ The coefficient of the (r + 1) th term from the end is 1 5 15 30 45 51 45 30 15
n 5 1
Cr . . . . . . . . . . . .
. . . . . . . . . . .
Hence, the coefficients of (r + 1) th term from the
i.e., The rows contains the coefficients for n = 0, 1, 2, 3,K.
beginning and the end are equal.
Each entry other than two entries at the ends is the sum of
(ii) The sum of the binomial coefficients in the three entries above it.
expansion of (1 + x ) n
15 = 1 + 4 + 10, 30 = 16 + 10 + 4, etc.
Q (1 + x ) n = n C 0 + n C 1 x + n C 2 x 2 + n C 3 x 3
Putting x = 1 and x = - 1 in Eq. (i), we get
+ ... + n C n x n
a 0 + a 1 + a 2 + a 3 + ... + a 2n = 3 n
Putting x = 1, we get
2 n = n C 0 + n C 1 + n C 2 + ... + n C n [sum of all coefficients] …(ii)
and a 0 - a 1 + a 2 - a 3 + ... + a 2n = 1 …(iii)
or C 0 + C 1 + C 2 + ... + C n = 2 n
On adding Eqs. (ii) and (iii), we get
\ Sum of binomial coefficients = 2 n
3n + 1
a 0 + a 2 + a 4 + ... + a 2n =
(iii) The sum of the coefficients of the odd terms 2
= The sum of the coefficients of the even terms [sum of coefficients of even powers of x ]
Q (1 + x ) n = n C 0 + n C 1 x + n C 2 x 2 + n C 3 x 3 On subtracting Eq. (ii) from Eq. (i), we get
+ ... + n C n x n
3n - 1
a 1 + a 3 + a 5 + K + a 2n - 1 =
Putting x = - 1, we get 2
0 = n C 0 - n C 1 + n C 2 - n C 3 + n C 4 - n C 5 + ... [sum of coefficients of odd powers of x ]
n
or C 1 + n C 3 + n C 5 + ... = n C 0 + n C 2 + n C 4 + ...
Since, the sum of all the coefficients is 2 n , therefore Putting x = i ( - 1 ) in Eq. (i), we get
2n a 0 + a 1 i + a 2 i 2 + a 3 i 3 + a 4 i 4 + a 5 i 5 + ... + a 2n i 2n = i n
each side is equal to i.e. 2 n -1 .
2 Þ (a 0 - a 2 + a 4 - ... ) + i (a 1 - a + a 5 - ... ) = i n
3
Hence, C 1 + C 3 + C 5 + ... = C 0 + C 2 + C 4 + ... = 2 n -1
(a 0 - a 2 + a 4 - K ) + i(a 1 - a 3 + a 5 - K )
n
Remark æ p pö æn p ö æn p ö
n
1. In the expansion of ( x - 2 y + 3z ) , putting x = y = z = 1, then
= ç cos + i sin ÷ = cos ç ÷ + i sin ç ÷
è 2 2ø è 2 ø è 2 ø
we get the sum of coefficients = ( 1 - 2 + 3) n = 2n .
2. In the expansion of ( 1 + x + x 2 ) n, putting x = 1, we get the sum On comparing real and imaginary parts, we get
of coefficients = ( 1 + 1 + 1) n = 3n. æ np ö
a 0 - a 2 + a 4 - K = cos ç ÷
è 2 ø
 Chap 06 Binomial Theorem 449

æ np ö Þ ( 4t + 1) ( t - 1) = 0
and a 1 - a 3 + a 5 - K = sin ç ÷
è 2 ø \ t = 1, t ¹ -
1
Þ 3 3 x / 2 = 1 = 30
4
Putting x = w and w2 (cube roots of unity) in Eq. (i), we get
3x
a 0 + a 1 w + a 2 w2 + a 3 w 3 + a 4 w 4 + ... = 0 …(iv) \ = 0 or x = 0
2
and a 0 + a 1 w2 + a 2 w 4 + a 3 w 6 + a 4 w 8 + ... = 0 …(v)
y Example 30. Find the values of
On adding Eqs. (ii), (iv) and (v) and then dividing by 3, we 1 1 1
get (i) + + + ...
(n - 1)! (n - 3)! 3! (n - 5)! 5!
a 0 + a 3 + a 6 + ... = 3 n - 1
1 1 1 1
Note
n -1
(ii) + + + ... +
(i) a1 + a4 + a7 + K = a2 + a5 + a8 + K = 3 12! 10! 2! 8 ! 4 ! 12!
(ii) a0 + a4 + a8 + K =
1ì n æ np ö ü Sol. (i) Q 1! = 1
í3 + 1 + 2cos ç ÷ ý
4î è 2 øþ
\ The given series can be written as
(iii) a1 + a5 + a9 + K =
1ì n æ np ö ü
í3 - 1 + 2sinç ÷ ý 1 1 1
4î è 2 øþ + + + ... …(i)
1ì np ü (n - 1)! 1! (n - 3)! 3! (n - 5)! 5!
(iv) a0 + a6 + a12 + K = í3n + 1 + 2n + 1 cos æç ö÷ ý
6î è 3 øþ Q Sum of values of each terms in factorial are equal.
2n 2n
r -1 i.e. (n - 1) + 1 = (n - 3) + 3 = (n - 5) + 5 = ... = n
(v) å r × ar = n × 3n (vi) å( -1) × r × ar = - n
r =1 r =1 From Eq. (i),
1 é n! n! n! ù
y Example 28. Find the sum of coefficients in the n! ê (n - 1)! 1! + (n - 3)! 3! + (n - 5)! 5! + ...ú
ë û
expansion of the binomial ( 5p - 4q )n , where n is a
1 n 2n - 1
positive integer. =( C 1 + n C 3 + n C 5 + ...) =
n! n!
Sol. Putting p = q =1 in (5p - 4q )n , the required sum of coeffi-
(ii) Q 0! = 1
cients = (5 - 4 )n = 1n = 1
\The given series can be written as
y Example 29. In the expansion of ( 3 - x / 4 + 3 5x / 4 )n , if 1
+
1
+
1
+ ... +
1
…(ii)
12!0! 10! 2! 8! 4 ! 0! 12!
the sum of binomial coefficients is 64 and the term
with the greatest binomial coefficient exceeds the third Q Sum of values of each terms in factorial are equal
by (n - 1), find the value of x. i.e., 12 + 0 = 10 + 2 = 8 + 4 = ... = 12
Sol. Given sum of the binomial coefficients in the expansion of 1 é 12! 12! 12! 12! ù
(3- x / 4 + 35 x / 4 )n = 64
From Eq. (ii),
12! ê 12! 0! + 10! 2! + 8! 4 ! + ... + 0! 12! ú
ë û
Then, putting 3- x / 4 = 35 x / 4 = 1 1 12 212 - 1 211
12 12 12
n n 6
= ( C0 + C2 + C 4 + ... + C 12 ) = =
\ (1 + 1) = 64 Þ 2 = 2 12! 12! 12!
\ n =6
y Example 31. Prove that the sum of the coefficients
We know that, middle term has the greatest binomial
in the expansion of (1 + x - 3x 2 ) 2163
coefficients. Here, n = 6
æn ö
is - 1.
\ Middle term = ç + 1÷ th term = 4th term = T 4 Sol. Putting x = 1 in (1 + x - 3x 2 )2163 , the required sum of
è2 ø
coefficients = (1 + 1 - 3)2163 = ( - 1)2163 = - 1
and given that T 4 = ( n - 1) + T 3
Þ T 3 + 1 = ( 6 - 1) + T 2 + 1 y Example 32. If the sum of the coefficients in the
Þ 6C 3 (3- x / 4 )3 (3 5 x / 4 )3 = 5 + 6C 2 (3- x / 4 )4 (3 5 x / 4 )2 expansion of (ax 2 - 2x + 1) 35 is equal to the sum of
Þ 20 × 3 3 x = 5 + 15 × 3 3 x / 2 the coefficients in the expansion of ( x - ay ) 35 , find the
Let 3 3x /2 = t value of a.
\ 20 t 2 = 5 + 15 t Sol. Given, sum of the coefficients in the expansion of
(ax 2 - 2x + 1)35
Þ 4t 2 - 3t - 1 = 0
450 Textbook of Algebra

= Sum of the coefficients in the expansion of ( x - ay )35 40

Putting x = y = 1, we get
Putting x = 1, we get 0 = åar
r =0
(a - 1)35 = (1 - a )35 or a 0 + a1 + a 2 + a 3 + a 4 + a 5 + ... + a 39 + a 40 = 0 …(ii)
Þ (a - 1)35 = - (a - 1)35 Putting x = - 1 in Eq. (i), we get
40
Þ 2 (a - 1)35 = 0 ( - 2)20 = å( - 1)r ar
Þ a -1 = 0 r =0

\ a =1 or a 0 - a1 + a 2 - a 3 + a 4 - a 5 + ... - a 39 + a 40 = 220 …(iii)

40 On subtracting Eq. (iii) from Eq. (ii), we get
y Example 33. If (1 + x - 2x 2 ) 20 = å ar x r , then find 2 [a1 + a 3 + a 5 + ... + a 39 ] = - 220
r =0
the value ofa1 + a 3 + a 5 + ... + a 39 . or a1 + a 3 + a 5 + ... + a 39 = - 219
40 Corollary On adding Eqs. (ii) and (iii) and then dividing by
Sol. Q (1 + x - 2x 2 )20 = åar x r …(i) 2, we get a 0 + a 2 + a 4 + ... + a 40 = 219
r =0

#L Exercise for Session 2

1. If the rth term in the expansion of (1 + x )20 has its coefficient equal to that of the (r + 4)th term, then r is
(a) 7 (b) 9 (c) 11 (d) 13
n
æ 1ö 5
2. If the fourth term in the expansion of ç px + ÷ is , then n + p is equal to
è xø 2
9 11 13 15
(a) (b) (c) (d)
2 2 2 2
n
æ 1ö 1
3. If in the expansion of ç 3 2 + 3 ÷ , the ratio of 7th term from the beginning to the 7th term from the end is ,
è 3ø 6
then n is
(a) 3 (b) 5 (c) 7 (d) 9

4. The number of integral terms in the expansion of (51/ 2 + 71/ 8 )1024 is

(a) 128 (b) 129 (c) 130 (d) 131

5. In the expansion of (7 1/ 3
+ 11 )1/ 9 6561
, the number of terms free from radicals is
(a) 715 (b) 725 (c) 730 (d) 750

6. n
If the coefficients of three consecutive terms in the expansion of (1 + x ) are 165, 330 and 462 respectively, the
value of n is
(a) 7 (b) 9 (c) 11 (d) 13

7. If the coefficients of 5th, 6th and 7th terms in the expansion of (1 + x )n are in AP, then n is equal to
(a) 7 only (b) 14 only (c) 7 or 14 (d) None of these
n
æ 1ö
8. If the middle term in the expansion of ç x 2 + ÷ is 924 x 6, the value of n is
è xø
(a) 8 (b) 12 (c) 16 (d) 20
n
æ 2 ö
9. If the sum of the binomial coefficients in the expansion of ç x 2 + 3 ÷ is 243, the term independent of x is equal to
è x ø
(a) 40 (b) 30 (c) 20 (d) 10

10. 2
In the expansion of (1 + x ) (1 + x + x ) ... (1 + x + x + ... + x 2 2n
), the sum of the coefficients is
(a) 1 (b) 2n ! (c) 2n ! + 1 (d) (2n + 1) !
Session 3
Two Important Theorems, Divisibility Problems
Two Important Theorems Now, let ( P - Q ) n = f ¢ , where 0 < f ¢ < 1
Theorem 1 If ( P + Q ) n = I + f , where I and n are Also, I + f = (P + Q )n …(i)
positive integers, n being odd and 0 £ f < 1 , then 0 £ f <1 …(ii)
show that ( I + f ) f = k n , where P - Q 2 = k > 0 and f ¢ = (P - Q )n …(iii)
P - Q < 1. and 0 < f ¢ <1 …(iv)
Proof Given, P -Q <1 \ 0 < ( P - Q )n < 1 On adding Eqs. (i) and (iii), we get
Now, let ( P - Q ) n = f ¢, where 0 < f ¢ < 1 I + f + f ¢ = (P + Q )n + (P - Q )n

Also I + f = ( P + Q )n …(i) = 2 [ n C 0 P n + n C 2 P n - 2 ( Q ) 2 + n C 4 P n - 4 ( Q ) 4 + ... ]

0 £ f <1 …(ii) = 2 (integer) = Even integer …(v)
f ¢ = ( P - Q )n …(iii) [Since, RHS contains even power of Q , so RHS is
an even integer]
and 0 < f ¢ <1 …(iv) \ LHS is also an integer.
On subtracting Eq. (iii) from Eq. (i), we get Q I is an integer.
I + f - f ¢ = ( P + Q )n - ( P - Q )n Þ f + f ¢ is also an integer.
\ f + f ¢ =1 [Q0 < ( f + f ¢ ) < 2 ]
= 2 [ n C 1 ( P ) n - 1 × Q + n C 3 ( P ) n - 3 × Q 3 + ... ]
or f ¢ =1- f
= 2 (integer) = Even integer …(v)
From Eq. (v), I = even integer - 1 = odd integer and
[Since, n is odd, RHS contains even powers
( I + f ) (1 - f ) = ( I + f ) f ¢
of P , so RHS is an even integer]
\ LHS is also an integer. = (P + Q )n (P - Q )n = (P 2 - Q )n = k n
Q I is an integer.
y Example 34. Show that the integral part of
\ (f - f ¢) is also an integer.
( 5 + 2 6 )n is odd, where n is natural number.
Þ f - f ¢ =0 [Q - 1 < ( f - f ¢ ) < 1]
or f =f¢ Sol. (5 + 2 6 )n can be written as (5 + 24 )n
From Eq. (v), I is an even integer and Now, let I + f = (5 + 24 )n …(i)
0£ f <1 …(ii)
(I + f ) f = (I + f ) f ¢ = ( P + Q )n ( P - Q )n
and let f ¢ = (5 - 24 )n …(iii)
= (P - Q 2 )n = k n
0< f ¢<1 …(iv)
Remark On adding Eqs. (i) and (iii), we get
If n is even integer, then ( P + Q) n + ( P - Q) n = I + f + f ¢
I + f + f ¢ = (5 + 24 )n + (5 - 24 )n
Since, LHS and I are integers.
\ (f + f ¢) is also an integer. I + 1 = 2p ,
Þ f + f¢ = 1 [Q0 < ( f + f ¢) < 2] " p Î N = Even integer [from theorem 2]
\ f¢ = 1- f \ I = 2p - 1 = Odd integer
Hence, ( I + f ) ( 1 - f ) = ( I + f ) f ¢ = ( P + Q) n ( P - Q) n
= ( P - Q2 ) n = k n y Example 35. Show that the integral part of
( 5 5 + 11) 2n +1 is even, where n Î N .
Theorem 2 If ( P + Q ) n = I + f , where I and n are
Sol. (5 5 + 11)2n + 1 can be written as ( 125 + 11)2n + 1
positive integers and 0 £ f < 1 , show that ( I + f )
(1 - f ) = k n , where P 2 - Q = k > 0 and P - Q < 1. Now, let I + f = ( 125 + 11)2 n + 1 …(i)

Proof Given, P - Q <1 0£ f <1 …(ii)

and let f ¢ = ( 125 - 11)2n + 1 …(iii)
\ 0 < (P - Q )n < 1
0< f ¢<1 …(iv)
452 Textbook of Algebra

On subtracting Eq. (iii) from Eq. (i), we get Now, let f ¢ = (8 - 63 )n …(iii)
I + f - f ¢ = ( 125 + 11)2 n + 1 - ( 125 - 11)2 n + 1 0< f ¢<1 …(iv)
I + 0 = 2p , " p Î N = Even integer On adding Eqs. (i) and (iii), we get
[from theorem 1] [ x ] + f + f ¢ = (8 + 63 )n + (8 - 63 )n
\ I = 2p = Even integer [ x ] + 1 = 2p , " p Î N = Even integer
[from theorem 2]
y Example 36. Let R = (6 6 + 14 ) 2n + 1 and f = R - [R ], \ [ x ] = 2p - 1 = Odd integer
where [ × ] denotes the greatest integer function. Find i.e., Integral part of x = Odd integer
the value of Rf , n Î N . Q f + f ¢=1 Þ 1- f = f ¢ …(v)
Sol. (6 6 + 14 )2 n + 1 can be written as ( 216 + 14 )2 n + 1 and LHS = x - x 2 + x [ x ] = x - x ( x - [ x ]) = x - xf
given that f = R - [ R ]
[Q x = [ x ] + f ]
and R = (6 6 + 14 )2 n + 1= ( 216 + 14 )2 n + 1 = x (1 - f ) = x f ¢ [from Eq.(v)]
\ [ R ] + f = ( 216 + 14 )2 n + 1 …(i) = (8 + 63 )n (8 - 63 )n [from Eqs.(i) and (iii)]
0£ f <1 …(ii)
= (64 - 63)n = (1)n = 1 = RHS
Let f ¢ = ( 216 - 14 )2 n + 1 …(iii)
0< f ¢<1 …(iv) Remark
On subtracting Eq. (iii) from Eq. (i), we get Sometimes, students find it difficult to decide whether a problem
2n + 1 2n + 1
is on addition or subtraction. Now, if x = [ x ] + f and 0 < f ¢ < 1
[ R ] + f - f ¢ = ( 216 + 14 ) - ( 216 - 14 ) and if [ x ] + f + f ¢= Integer. Then, addition and if
[ R ] + 0 = 2p , " p Î N = Even integer [from theorem 1] [ x ] + f - f ¢ = Integer, the subtraction and values of (f + f ¢) and
(f - f ¢) are 1and 0, respectively.
\ f - f ¢ = 0 or f = f ¢
Now, Rf = Rf ¢ = ( 216 + 14 )2 n + 1 ( 216 - 14 )2 n + 1
= (216 - 196)2 n + 1 = (20)2 n + 1 Divisibility Problems
y Example 37. If (7 + 4 3 )n = s + t, where n and s are Type I
positive integers and t is a proper fraction, show that (i) ( x n - a n ) is divisible by ( x - a ), " n Î N .
(1 - t ) (s + t ) = 1. (ii) ( x n + a n ) is divisible by ( x + a ), " n ÎOnly odd
Sol. (7 + 4 3 )n can be written as (7 + 48 )n natural numbers.
\ s + t = (7 + 48 )n …(i)
y Example 39. Show that
0<t <1 …(ii)
Now, let t ¢ = (7 - 48 )n …(iii) 19921998 - 19551998 - 1938 1998 + 19011998 is divisible by 1998.
0<t¢<1 …(iv) Sol. Here, n = 1998 (Even)
On adding Eqs. (i) and (iii), we get \ Only result (i) applicable.
s + t + t ¢ = (7 + 48 )n + (7 - 48 )n Let P = 19921998 - 19551998 - 19381998 + 19011998
s + 1 = 2p , " p Î N = Even integer [from theorem 2] = (19921998 - 19551998 ) - (19381998 - 19011998 )
\ t + t ¢ = 1 or 1 - t = t ¢ divisible by (1992 - 1955 ) divisible by (1938 - 1901)
Then, (1 - t ) (s + t ) = t ¢ (s + t ) = (7 - 48 )n (7 + 48 )n i. e. 37 i. e. 37
[from Eqs. (i) and (iii)] \ P is divisible by 37.
= ( 49 - 48) = (1)n = 1
n
Also, P = (19921998 - 19381998 ) - (19551998 - 19011998 )
divisible by (1992 - 1938 ) divisible by (1955 - 1901)
y Example 38. If x = (8 + 3 7 )n , where n is a natural i.e., 54 i.e., 54
number, prove that the integral part of x is an odd \ P is also divisible by 54.
integer and also show that x - x 2 + x [ x ] = 1 , where [ × ] Hence, P is divisible by 37 ´ 54, i.e., 1998.
denotes the greatest integer function. y Example 40. Prove that 2222 5555 + 5555 2222 is
Sol. (8 + 3 7 )n can be written as (8 + 63 )n
divisible by 7.
\ x = [x ] + f
Sol. We have, 22225555 + 55552222
or [ x ] + f = (8 + 63 )n …(i)
0£ f <1 …(ii) = (22225555 + 4 5555 ) + (55552222 - 4 2222 ) - ( 4 5555 - 4 2222 ) …(i)
 Chap 06 Binomial Theorem 453

The number (22225555 + 4 5555 ) is divisible by 2222 + 4

= 2226 = 7 ´ 318, which is divisible by 7 and the number
How to Find Remainder
(55552222 - 4 2222 ) is divisible by by Using Binomial Theorem
5555 - 4 = 5551 = 7 ´ 793, which is divisible by 7 and the If a, p, n and r are positive integers, then to find the
number remainder when a pn + r is divided by b, we adjust power of
( 4 5555 - 4 2222 ) = 4 2222 ( 4 3333 - 1) = 4 2222 (641111 - 11111 ) is a to a pn + r which is very close to b, say with difference 1
divisible by 64 - 1 = 63 = 7 ´ 9, which is divisible by 7. i.e., b ± 1. Also, the remainder is always positive. When
Therefore, each brackets of Eq. (i) are divisible by 7. Hence, number of the type 5n - 2 is divided by 5, then we have
22225555 + 55552222 is divisible by 7. 5 ) 5n - 2 (n
5n
Type II To show that an Expression -
-2
is Divisible by An Integer We can write - 2 = - 2 - 3 + 3 = - 5 + 3
Solution Process 5n - 2 5n - 5 + 3 3
or = =n -1+
(i) If a, p, n and r are positive integers, first of all write 5 5 5
a pn + r = a pn × a r = (a p ) n × a r Hence, the remainder is 3.
(ii) If we will show that the given expression is divisible y Example 43. If 7 103 is divided by 25, find the
by c. Then, expression a p = {1 + (a p - 1)}, if some
remainder.
power of (a p - 1) has c as a factor. Soln. We have, 7103 = 7 × 7102 = 7 × (7 2 )51 = 7 (49)51 = 7 (50 - 1)51
or a p = {2 + (a p - 2 )}, if some power of (a p - 2 ) has c = 7 [(50)51 - 51C 1 (50)50 + 51C 2 (50)49 - ... - 1]
as a factor.
= 7 [(50)51 - 51C 1 (50)50 + 51C 2 (50)49 - ... + 51C 50 (50)]
or a p = {3 + (a p - 3 )}, if some power of (a p - 3 ) has c
- 7 - 18 + 18
as a factor.
M M M M M M = 7 [ 50 ((50)50 - 51 C 1(50)49 + 51 C 2 (50)48 -...+ 51C 50 )] - 25 + 18
or a p = {k + (a p - k )}, if some power of (a p - k ) has c = 7 [ 50k ] - 25 + 18 , where k is an integer.
= 25 [14k - 1] + 18 = 25p + 18 [where p is an integer]
as a factor.
7103 18
Now, = p + . Hence, the remainder is 18.
y Example 41. If n is any positive integer, show 25 25
that 2 3n + 3 - 7n - 8 is divisible by 49. 5 ...
5

Sol. Given expression y Example 44. Find the remainder, when 5 5

= 23n + 3 - 7n - 8 = 23n × 23 - 7n - 8 ( 24 times 5) is divided by 24.
5
= 8n × 8 - 7n - 8 = 8 (1 + 7 )n - 7n - 8 Sol. Here, 555
5 ...
(23 times 5) is an odd natural number.
= 8 (1 + n C 1 × 7 + n C 2 × 7 2 + ... + n C n × 7n ) - 7n - 8 5 5 ...
Let 5 (23 times 5) = 2m + 1
= 8 + 56n + 8 (n C 2 × 7 2 + ... + n C n × 7n ) - 7n - 8 5
5...
= 49n + 8 (n C 2 × 7 2 + ... + n C n × 7n ) Now, let x = 55 (24 times 5) = 52 m + 1 = 5 × 52 m , where m is
= 49 {n + 8 (n C 2 + ... + n C n × 7n - 2 )} a natural number.
\ x = 5 × (52 )m = 5 ( 24 + 1)m
Hence, 23n + 3 - 7n - 8 is divisible by 49. = 5 [ mC 0 (24 )m + mC 1 (24 )m - 1 + ... + mC m - 1 (24 ) + 1]
y Example 42. If 10n divides the number 101100 - 1, find = 5 (24k + 1) = 24 (5k ) + 5
x 5
the greatest value of n. \ = 5k +
24 24
Sol. We have, 101100 - 1 = (1 + 100)100 - 1 Hence, the remainder is 5.
100 100
=1+ C 1 × 100 + C 2 × 1002 + ... + 100
C 100 100100 - 1 32
= 100
C 1 × 100 + 100 2
C 2 × 100 + K + 100
C 100 × 10100
y Example 45. If 7 divides 32 32 , then find the remainder.
= (100) (100) + 100
C 2 × 1002 + ... + 100
C 100 × 100100 Solution. We have, 32 = 25
= (100)2 [1 + 100
C 2 + ... + 10098 ] \ 3232 = (25 )32 = 2160 = (3 - 1)160
160
= 1002 k, where k is a positive integer = C 0 (3)160 - 160
C 1 (3)159 + ... - 160
C 159 (3) + 1
159 160 158 160
Therefore, 101100 - 1 is divisible by 1002 i.e., 104 . = 3 (3 - C 1 ( 3) + ... - C 159 ) + 1
+
\ n=4 = 3m + 1, m Î I
454 Textbook of Algebra

32
Now, 3232 = 32 3m + 1 = 2 5 (3m + 1) = 215m + 5 (i) For last digit
2 3 ( 5m + 1) 5m + 1 5m + 1
= 2 ×2 = 4 ( 8) = 4 ( 7 + 1) 17 256 = 290[ 128 C 0 (290)127 - 128
C 1(290)126
5m + 1
= 4[ C 0 (7 )5m + 1 + 5m + 1
C 1 (7 )5m + 5m + 1C 2 (7 )5m - 1 + 128
C 2 (290)125 - ... -128 C 127 (1)] + 1
+ ... + 5m + 1C 5m (7 ) + 1]
= 290 (k ) + 1, where k is an integer.
= 4 [ 7 ( 5m + 1C 0 (7 )5m + 5m + 1C 1(7 )5m - 1 +... + 5m + 1C 5m ) + 1]
= 4 [7k + 1] , where k is positive integer = 28k + 4 \ Last digit = 0 + 1 = 1
32 (ii) For last two digits,
3232 4
\ = 4k +
7 7 17 256 = (290)2 [ 128C 0 (290)126 - 128
C 1 (290)125 +
Hence, the remainder is 4. 128
C 2 (290)124 - ... + 128
C 126 (1)] - 128
C 127 (290) + 1
128
= 100 m - C 127 (290) + 1, where m is an integer.
How to Find Last Digit, Last Two Digits, 128
= 100 m - C 1 (290) + 1 = 100 m - 128 ´ 290 + 1
Last Three Digits, ... and so on.
= 100 m - 128 ´ (300 - 10) + 1
Ifa, p, n andr are positive integers, thena pn + r is adjust of the
= 100 (m - 384 ) + 1281
form(10k ± 1) m , wherek andm are positive integers. For last = 100 n + 1281, where n is an integer.
digit, take 10 common. For last two digits, take 100 common, \ Last two digits = 00 + 81 = 81
for last three digits, take 1000 common , ... and so on.
(iii) For last three digits,
i.e. (10k ± 1) m = (10k ) m + m C 1 (10k ) m - 1 ( ± 1)
17 256 = (290)3 [ 128C 0 (290)125 - 128
C 1 (290)124
+ m C 2 (10k ) m - 2 ( ± 1) 2 + ... +
128
m
C m - 2 (10k ) 2 ( ± 1) m -2 + m C m - 1 (10k ) ( ± 1) m - 1 + ( ± 1) m + C 2 (290)123 - ... - 128
C 125 (1)]
128
+ C 126 (290)2 - 128
C 127 (290) + 1
For last digit = 10 l + ( ± 1) m
128 2 128
= 1000 m + C 126 (290) - C 127 (290) + 1
For last two digits = 100 m + m C m - 1 (10k ) ( ± 1) m - 1 + ( ± 1) m
where, m is an integer
For last three digits = 1000 n +m C m -2 (10k ) 2 ( ±1) m -2 +m C m -1 128
= 1000 m + C 2 (290)2 - 128
C 1 (290) + 1
m -1 m
(10k ) ( ± 1) + ( ± 1) and so on where l, m, n Î I.
(128) (127 )
= 1000 m + (290)2 - 128 ´ 290 + 1
y Example 46. Find the last two digits of 3 400
. 2
400 2 200 200 200
= 1000 m + (128) (127 ) (290) (145) - (128) (290) + 1
Sol. We have, 3 = (3 ) = (9) = (10 - 1)
= 1000 m + (128) (290) (127 ´ 145 - 1) + 1
200 200 199
= (10) - C 1 (10) + 200C 2 (10)198 - 200
C 3 (10)197
= 1000 m + (128) (290) (18414 ) + 1
200
+ ... + C 198 (10)2 - 200
C 199 (10) + 1 = 1000 m + 683527680 + 1
200
= 100 m - C 199 (10) + 1, where m Î I = 1000 m + 683527000 + 680 + 1
= 100 m - 200
C 1 (10) + 1 = 100 m - 2000 + 1 = 1000 (m + 683527 ) + 681
= 100 (m - 20) + 1 = 100 p + 1, where p is an integer. \ Last three digits = 000 + 681 = 681

Hence, the last two digits of 3400 is 00 + 1 = 01.

Two Important Results
y Example 47. If the number is 17 256 , find the n
æ 1ö
(i) last digit. (ii) last two digits. (i) 2 £ ç 1 + ÷ < 3, n ³ 1, n Î N
è nø n n
(iii) last three digits of 17 256 . ænö ænö
(ii) If n > 6, then ç ÷ < n ! < ç ÷
è3ø è2ø
Sol. Since, 17 256 = (17 2 )128 = (289 )128 = (290 - 1)128

\ 17 256 = 128C 0 (290)128 - 128

C 1 (290)127 + 128
C 2 (290)126
y Example 48. Find the positive integer just greater
128
than (1 + 00001
. )10000 .
- C 3 (290)125 + ... - 128
C 125 (290)3 + 128
C 126 (290)2 10000
æ 1 ö
128 Sol. (1 + 0.0001)10000 = ç1 + ÷
- C 127 (290) + 1 è 10000 ø
 Chap 06 Binomial Theorem 455

n
æ 1ö y Example 50. Find the greater number in 300! and
We know that, 2 £ ç1 + ÷ < 3, n ³ 1, n Î N [Result (i)]
è nø
300 300 .
Hence, positive integer just greater than (1 + 0.0001)10000
is 3.
Sol. Since, (100)150 > 3150
y Example 49. Find the greater number is 100100 and Þ (100)150 × (100)150 > 3150 × (100)150
( 300)! . Þ (100)300 > (300)150
n
æn ö (100)300 > 300300
Sol. Using Result (ii), We know that, ç ÷ < n ! or …(i)
è3ø n
Putting n = 300, we get æn ö
Using result (ii), ç ÷ < n !
(100)300 < (300) ! …(i) è3ø

But (100) 100

< (100) 300
…(ii) Putting n = 300, we get (100)300 < 300 ! …(ii)
From Eqs. (i) and (ii), we get From Eqs. (i) and (ii), we get
(100)100 < (100)300 < (300) ! 300 300 < (100)300 < 300 !
Þ (100)100 < (300) ! Þ 300 300 < 300 !
Hence, the greater number is (300) ! . Hence, the greater number is 300 !.

#L Exercise for Session 3

1. If x = (7 + 4 3 )2n = [ x ] + f , where n Î N and 0 £ f < 1, then x (1 - f ) is equal to
(a) 1 (b) 0 (c) - 1 (d) even integer

2. If (5 + 2 6 )n = I + f ; n, I Î N and 0 £ f < 1, then I equals

1 1 1 1
(a) -f (b) -f (c) -f (d) +f
f 1+ f 1- f 1+ f

1- f 2
3. If n > 0 is an odd integer and x = ( 2 + 1)n, f = x - [ x ], then is
f
(a) an irrational number (b) a non-integer rational number (c) an odd number (d) an even number

4. Integral part of ( 2 + 1)6 is

(a) 196 (b) 197 (c) 198 (d) 199

5. (103) 86
- (86) 103
is divisible by
(a) 7 (b) 13 (c) 17 (d) 23
78
2
6. Fractional part of is
31
2 4 8 16
(a) (b) (c) (d)
31 31 31 31

7. The unit digit of 171983 + 111983 - 71983 is

(a) 1 (b) 2 (c) 3 (d) 0

8. The last two digits of the number (23) 14

are
(a) 01 (b) 03 (c) 09 (d) 27

9. The last four digits of the number 3100 are

(a) 2001 (b) 3211 (c) 1231 (d) 0001

10. The remainder when 2323 is divided by 53 is

(a) 17 (b) 21 (c) 30 (d) 47
Session 4
Use of Complex Numbers in Binomial Theorem, Multinomial
Theorem, Use of Differentiation, Use of Integration, Binomial
Inside Binomial, Sum of the Series

Use of Complex Numbers From Eqs. (i) and (ii), we get

in Binomial Theorem (C 0 - C 2 + C 4 - ...) + i (C 1 - C 3 + C 5 - ...)

æ np ö
= 2 n / 2 cos ç ÷ + i ×2 n / 2 sin ç ÷
æ np ö
If q Î R, n Î N and i = - 1, then è 4 ø è 4 ø
On comparing real and imaginary parts, we get
(cos q + i sin q ) n = n C 0 (cos q ) n - 0 (i sin q ) 0
æ np ö
+ n C 1 (cos q ) n - 1 (i sin q ) 1 C 0 - C 2 + C 4 - ... = 2 n / 2 cos ç ÷ [part (i)]
è 4 ø
n n -2 2 n n-3
+ C 2 (cos q ) (i sin q ) + C 3 (cos q )
æ np ö
3 C 1 - C 3 + C 5 - ... = 2 n / 2 sin ç ÷ [part (ii)]
(i sin q ) + ... è 4 ø
or cos nq + i sin nq = cos n q + i × n C 1 (cos q ) n - 1 sin q We have, (1 + x )n = C 0 + C 1x + C 2 x 2 + C 3 x 3
+ C4 x 4

- n C 2 (cos q ) n - 2 sin2 q - i × n C 3 (cos q ) n - 3 sin 3 q +... + C5 x 5

+ C6 x 6
+ ...
On comparing real and imaginary parts, we get 2
Putting x = 1, w, w (cube roots of unity) and adding, we get
cos nq = cos n q - n C 2 (cos q ) n - 2 sin2 q 3 ( C 0 + C 3 + C 6 + ...) = 2n + (1 + w)n + (1 + w2 )n
n n-4 4
- C 4 (cos q ) sin q - ... = 2n + ( - w2 )n + ( - w)n = 2n + ( - 1)n ( w2n + wn )
n n -1 n n-3 3
andsin nq = C 1 (cos q ) sin q - C 3 (cos q ) sin q ìï 4 pin 2 pin ü
n n ï
= 2 + ( -1) íe 3 +e 3 ý
+ n C 5 (cos q ) n - 5 sin5 q - ... ïî ïþ
y Example 51. If (1 + x )n = C 0 + C 1 x + C 2 x 2
æ np ö
= 2n + ( - 1)n × e npi × 2 cos ç ÷
+ C3 x 3 + C4 x 4
+ ... , find the values of è 3 ø
æ np ö
(i) C 0 - C 2 + C 4 - C 6 + ... = 2n + ( - 1)n × ( - 1)n × 2 cos ç ÷
è 3 ø
(ii) C 1 - C 3 + C 5 - C 7 + ...
æ np ö æ np ö
(iii) C 0 + C 3 + C 6 + ... = 2 n + ( - 1)2 n × 2 cos ç ÷ = 2n + 2 cos ç ÷
è 3 ø è 3 ø
Sol. Q(1 + x )n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + C 4 x 4
1ì n æ np öü
\ C 0 + C 3 + C 6 + ... = í2 + 2 cos ç ÷ý
+ C 5 x 5 + ... 3î è 3 øþ
Putting x = i , where i = - 1, then
y Example 52. Find the value of
(1 + i )n = C 0 + C 1 i + C 2 i 2 + C 3 i 3 + C 4 i 4 + C 5 i 5 + ... 4n
C 0 + 4n C 4 + 4n C 8 + ... + 4n
C 4n .
= (C 0 - C 2 + C 4 - ...) + i (C 1 - C 3 + C 5 - ...) …(i)
Sol. Q 4 - 0 = 8 - 4 = ... = 4
n
é æ 1 i öù
Also, (1 + i )n = ê 2 ç + ÷ \ Four roots of unity (1)1/ 4 are 1, - 1, i , - i , we have
ë è 2 2 ø úû
n
(1 + x )4n = 4n
C0 + 4n
C 1x + 4n
C 2x 2 + 4n
C 3 x 3 + ...
æ p pö
= 2n / 2 ç cos + i sin ÷ Putting x = 1, - 1, i , - i and then adding, we get
è 4 4ø
4 ( 4n C 0 + 4n
C 4 + 4n C 8 + ...) = 24n + 0 + (1 + i )4n + (1 - i )4n
n/2 æ np np ö
=2 ç cos + i sin ÷ …(ii) = 24n + (2i )2 n + ( - 2i )2 n
è 4 4 ø
 Chap 06 Binomial Theorem 457

= 24n + 22 n ( - 1)n + 22 n ( - 1)n y +z =5

=2 4n
+ ( - 1) × 2 n 2n + 1 On adding all, we get 2 ( x + y + z ) = 12
\ x +y +z =6
\ 4n
C0 + 4n
C4 + 4n
C 8 + ... = 24n - 2 + ( - 1)n × 22n - 1
Then, x = 1, y = 3, z = 2
Remark Therefore, the coefficient of a 3b 4c 5 in the expansion of
If ( 1 + x ) n = C0 + C1 x + C2 x 2 + C3 x 3+ K + Cn x n , then (bc + ca + ab )6 or the coefficient of (ab )1 (bc )3 (ca )2 in the
np 6!
(i) C0 + C4 + C8 + C12 + K =
1ì n -1
+ 2n / 2cos æç ö÷ üý expansion of (bc + ca + ab )6 is , i.e. 60.
í2 è 4 øþ 1!3!2!
2î
1ì n -1 np
+ 2n / 2sin æç ö÷ üý
(ii) C1 + C5 + C9 + C13 + K =
Aliter
í2 è 4 øþ
2î Coefficient of a 3b 4c 5 in the expansion of (bc + ca + ab )6
np
(iii) C0 + C6 + C12 + K = ìí2n - 1 cos æç ö÷ + 3n / 2cos æç ö÷ üý
1 np
3î è4 ø è 6 øþ = Coefficient of a 3b 4c 5 in the
6
æ1 1 1ö
expansion of (abc )6 ç + + ÷
èa b c ø
Multinomial Theorem 3
æ1ö æ1ö æ1ö
2 1
= Coefficient of ç ÷ ç ÷ ç ÷ in the expansion of
If n is a positive integer and x 1 , x 2 , x 3 , ... , x k Î C , then èa ø èb ø èc ø
( x 1 + x 2 + x 3 + ... + x k ) n = S
n! æ1 1 1ö 6!
6

(a 1 !)(a 2 !)(a 3 !) ... (a k !) ç + + ÷ is = 60

èa b c ø 3 ! 2 ! 1 !
a a a
x 1 1 x 2 2 x 3 3 ... x k a k
where, a 1 , a 2 , a 3 , ... , a k are all non-negative integers
such that a 1 + a 2 + a 3 + ... + a k = n.
Number of Distinct or Dissimilar
Terms in the Multinomial Expansion
Remark Statement The number of distinct or dissimilar terms in
The coefficient of x1a1 × x 2a 2 × x 3a 3 ... x k a k in the expansion of
the multinomial expansion of ( x 1 + x 2 + x 3 + ... + x k ) n
( x1 + x 2 + x 3 + ... + x k ) n is S
n!
.
( a1 !) ( a2 !) ( a3 !) ... ( ak !) n + k -1
is Ck - 1 .
In Particular
Proof We have, ( x 1 + x 2 + x 3 + ... + x k ) n
(i) (a + b + c ) = S
n!
n
a a b b c g such that
=S
n! a a a a
(a !) (b !) ( g !) x 1 1 x 2 2 x 3 3 ... x k k
(a 1 !) (a 2 !) (a 3 !) ... (a k !)
a +b + g =n
where, a 1 , a 2 , a 3 , ... , a k are non-negative integers such
(ii) (a + b + c + d ) n = S
n!
a a bb c g d d that
(a !) (b !) ( g !) ( d !)
such that a + b + g + d = n a 1 + a 2 + a 3 + ... + a k = n ...(i)
Here, the number of terms in the expansion of
y Example 53. Find the coefficient of a 4 b 3 c 2 d in (x 1 + x 2 + x 3 + K + x k )n
the expansion of (a - b + c - d ) 10. = The number of non-negative integral solutions of the Eq. (i)
4 3 2
Sol. The coefficient of a b c d in the expansion of = n + k - 1C k - 1
10 !
(a - b + c - d )10 is ( - 1)4 = 12600 y Example 55. Find the total number of distinct or
4 !3!2!1!
dissimilar terms in the expansion of
[powers of b and d are 3 and 1 \( -1)3 ( -1) ]
( x + y + z + w )n , n Î N .
y Example 54. Find the coefficient of a 3b 4 c 5
in the Sol. The total number of distinct or dissimilar terms in the
6 expansion of ( x + y + z + w )n is
expansion of (bc + ca + ab ) .
( n + 3) ( n + 2) ( n + 1)
Sol. In this case, write a 3b 4c 5
= (ab )x (bc )y (ca )z say = n + 4 - 1C 4 - 1 = n + 3C 3 =
1×2×3
\ a 3b 4c 5
= a z + x ×b x +y
×c y +z
( n + 1) ( n + 2) ( n + 3)
=
Þ z + x = 3, x + y = 4 6
458 Textbook of Algebra

I. Aliter
We know that, ( x + y + z + w )n = {( x + y ) + (z + w )}n
Coefficient of x r in
= ( x + y )n + n C 1 ( x + y )n - 1 (z + w ) Multinomial Expansion
+ n C 2 ( x + y )n - 2 (z + w )2 + ... + n C n (z + w )n If n is a positive integer and a 1 , a 2 , a 3 , ... , a k Î C , then
\ Number of terms in RHS coefficient of x r in the expansion of (a 1 + a 2 x + a 3 x 2
= (n + 1) + n × 2 + (n - 1) × 3 + ... + 1 × (n + 1) k -1 n
+ ... + a k x ) , is
n
= S
r =0
( n - r + 1) ( r + 1)
S n! a a2 a a
a 1 1 a 2 a 3 3 ... a k k
n n n n (a 1 !) (a 2 !) (a 3 !) ... (a k !)
= S (n + 1) + nr - r 2 = (n + 1)r S= 0 1 + n r S= 0 r - r S= 0 r 2
r =0 where, a 1 , a 2 , a 3 , ... , a k are non-negative integers such
n ( n + 1) n ( n + 1) ( 2 n + 1) that a 1 + a 2 + a 3 + ... + a k = n
= ( n + 1) × ( n + 1) + n × -
2 6 and a 2 + 2 a 3 + 3 a 4 + ... + (k - 1) a k = r
( n + 1) ( n + 2) ( n + 3)
= y Example 57. Find the coefficient of x 7 in the
6
II. Aliter expansion of (1 + 3x - 2x 3 )10 .
( x + y + z + w )n = S
n!
x n1 y n 2 z n 3 w n 4 Sol. Coefficient of x 7 in the expansion of (1 + 3x - 2x 3 )10 is
n1 ! n 2 ! n 3 ! n 4 !
= S 10!
(1)a (3)b ( -2)g
where, n1, n 2 , n 3 ,n 4 are non-negative integers subject to the a ! b! g !
condition n1 + n 2 + n 3 + n 4 = n
where, a + b + g = 10 andb + 3g = 7
Hence, number of the distinct terms
The possible values of a , b and g are given below
= Coefficient of x n in ( x 0 + x 1 + x 2 + ... + x n )4
4 a b g
æ1 - x n + 1 ö
= Coefficient of x n in ç ÷ 3 7 0
è 1- x ø
5 4 1
= Coefficient of x n in (1 - x n + 1 )4 (1 - x )- 4
7 1 2
= Coefficient of x n in (1 - x )- 4 [Q x n + 1 > x n ]
7
( n + 3) ( n + 2) ( n + 1) \ Coefficient of x
= n + 3C n = n + 3C 3 =
6 10 ! 10 !
= ( 1) 3 ( 3) 7 ( - 2) 0 + (1)5 (3)4 ( - 2)1
3!7 !0! 5! 4 !1!
Greatest Coefficient in +
10 !
(1)7 (3)1 ( - 2)2
7 !1!2!
Multinomial Expansion = 262440 - 204120 + 4320 = 62640
The greatest coefficient in the expansion of
n!
( x 1 + x 2 + x 3 + ... + x k ) n is
(q !) k -r
((q + 1) !) r
, where q is
Use of Differentiation
the quotient and r is the remainder when n is divided by k i.e. This method applied only when the numericals occur as
k ) n (q the product of the binomial coefficients, if
r (1 + x ) n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + ... + C n x n
y Example 56. Find the greatest coefficient in the
expansion of (a + b + c + d )15 .
Solution Process
(i) If last term of the series leaving the plus or minus
Sol.Here, n = 15 and k = 4 [Qa, b, c , d are four terms] sign is m, then divide m by n. If q is the quotient and
4 ) 15 (3 r is the remainder.
12 i.e. m = nq + r or n ) m (q
3 nq
\ q = 3 and r = 3 r
15 ! q
Hence, greatest coefficient = Then, replace x by x in the given series and
(3!)1 ( 4 !) 3 multiplying both sides of the expression by x r .
 Chap 06 Binomial Theorem 459

(ii) After this, differentiate both sides w.r.t. x and put y Example 59. If (1 + x )n = C 0 + C 1 x + C 2 x 2

x = 1 or - 1 or i(i = -1 ), etc. According to the given

+ ... + C n x n , prove that
series.
(iii) If product of two numericals (or square of numericals) C 0 + 2 C 1 + 3 C 2 + ... + (n + 1) C n = (n + 2) 2n - 1 .
or three numericals (or cube of numericals), then Sol. Here, last term of C 0 + 2 C 1 + 3C 2 + ... + (n + 1) C n is
differentiate twice or thrice. (n + 1) C n i.e., (n + 1) and last term with positive sign.

y Example 58. If and n + 1 = n ×1 + 1

or n )n + 1 (1
(1 + x ) n = C 0 + C 1 x + C 2 x 2 + ... + C n x n , prove that
-n
C 1 + 2C 2 + 3C 3 + ... + nC n = n × 2n -1 . Here, q = 1 and r = 1 1
Sol. Here, last term of C 1 + 2C 2 + 3C 3 + ... + n C n is nC n i.e., n The given series is
and last term with positive sign. (1 + x )n = C 0 + C 1 x + C 2 x 2
+ ... + C n x n
Then, n = n × 1 + 0 or n ) n (1
Now, replacing x by x 1 and multiplying both sides by x, we
n
get
0
x (1 + x )n = C 0 x + C 1 x 2
+ C2 x 3
+ ... + C n x n + 1
Here, q = 1 and r = 0
Differentiating both sides w.r.t. x, we get
Then, the given series is
2 3 x × n (1 + x )n -1 + (1 + x )n × 1 = C 0 + 2C 1 x + 3C 2 x 2
(1 + x )n = C 0 + C 1 x + C 2 x + C3 x + ... + C n x n
Differentiating both sides w.r.t. x , we get + ... + (n + 1) C n x n
n (1 + x )n - 1 = 0 + C 1 + 2C 2 x + 3C 3 x 2 + ... + nC n x n - 1 Putting x = 1, we get
Putting x = 1, we get n (2)n - 1 + 2n = C 0 + 2C 1 + 3C 2 + K + (n + 1) C n
n × 2n - 1 = C 1 + 2C 2 + 3C 3 + ... + nC n or C 0 + 2C 1 + 3C 2 + ... + (n + 1) C n = (n + 2) 2n - 1
or C 1 + 2C 2 + 3C 3 + ... + nC n = n × 2n - 1 I. Aliter
I. Aliter LHS = C 0 + 2 C 1 + 3 C 2 + ... + (n + 1)C n
C 1 + 2C 2 + 3C 3 + ... + n C n = C 0 + (1 + 1) C 1 + (1 + 2) C 2 + ... + (1 + n ) C n
n ( n - 1) n ( n - 1) ( n - 2) = (C 0 + C 1 + C 2 + ... + C n ) + (C 1 + 2C 2 + ... + n C n )
= n +2 × +3× + ... + n × 1
1 ×2 1 ×2 ×3 [use example 58]
ì ( n - 1) ( n - 2) ü n n -1 n -1
= n í1 + (n - 1) + + ... + 1ý = 2 + n ×2 = ( n + 2) 2 = RHS
î 1 ×2 þ
II. Aliter
Let n - 1 = N , then
LHS = C 0 + 2 C 1 + 3 C 2 + ... + (n + 1) C n
ì N ( N - 1) ü n +1 n +1
LHS = (1 + N ) í1 + N + + ... + 1ý
î 1 ×2 þ = S r × nC r - 1 =rS= 1
r =1
( r - 1 + 1) × n C r -1
N N N
= (1 + N ) {1 + C 1 + C 2 + ... + C N } n +1

= (1 + N ) 2 N = n × 2n - 1 = RHS = S
r =1
( r - 1) × n C r -1 + nC r -1

II. Aliter n +1 n +1
n
= S n × n - 1C r - 2 + rS= 1 n
Cr
LHS = C 1 + 2 C 2 + 3 C 3 + ... + n C n = S r × nC r
r =1
r =1
-1

é n n n -1 ù
n
é n ù êQ C r = . Cr - 2ú
= S r × × n - 1C r
n n n -1 -1
-1 êëQ C r = r × Cr - 1ú ë r -1 û
r =1 r û
n -1 n -1 n -1 n -1
n = n (0 + C0 + C1 + C 2 + ... + Cn - 1)
=n S n - 1C r -1
r =1 + (n C 0 + n C 1 + n C 2 + ... + n C n )
n -1 n -1
=n( C0 + C 1 + n -1C 2 + ... + n -1
C n - 1] = n × 2n - 1 + 2n = (n + 2) × 2n - 1 = RHS
n -1
= n ×2 = RHS
460 Textbook of Algebra

y Example 60. If (1 + x )n = C 0 + C 1 x + C 2 x 2
y Example 61. If (1 + x )n = C 0 + C 1 x + C 2 x 2

+ ... + C n x n , prove that + ... + C n x n , prove that + 3 2 × C 3 + ... + n 2 × C n

C 0 + 3C 1 + 5C 2 + ... + (2n + 1) C n = (n + 1) 2n . 1 2 × C 1 + 2 2 × C 2 = n (n + 1) × 2n -2 .
Sol. Here, last term of C 0 + 3C 1 + 5C 2 + ... + (2n + 1) C n is Sol. Here, last term of 12 × C 1 + 22 × C 2 + 32 × C 3 + ... + n 2 × C n is
(2n + 1) C n i.e., (2n + 1) and last term with positive sign. n 2 × C n i.e., n 2 . Linear factors of n 2 are n and n; [start
Then, 2n + 1 = n × 2 + 1 always with greater factor] and last term with positive
sign.
or n ) 2n + 1(2
and n = n × 1 + 0 or n ) n (1
- 2n
1 -n
Here, q = 2 and r = 1 0
The given series is Here, q = 1 and r = 0
2 Then, the given series is
(1 + x )n = C 0 + C 1x + C 2 x + ... + C n x n
(1 + x )n = C 0 + C 1x + C 2 x 2 + C 3 x 3 + ... + C n x n
Now, replacing x by x 2 , we get
On differentiating both sides w.r.t. x, we get
(1 + x 2 )n = C 0 + C 1x 2
+ C2 x 4
+ ... + C n x 2n
nx (1 + x )n - 1 = C 1 + 2C 2 x + 3C 3 x 2
+ ... + n C n x n - 1 …(i)
On multiplying both sides by x 1, we get
and in last term, numerical is n C n i.e., n and power of
2n +1
x (1 + x 2 )n = C 0 x + C 1x 3 + C 2 x 5 + ... + C n x (1 + x ) is n - 1.
On differentiating both sides w.r.t. x, we get Then, n = (n - 1) × 1 + 1 or n - 1) n (1
x × n (1 + x 2 )n - 1 × 2x + (1 + x 2 )n × 1 = C 0 + 3C 1 x 2
+ 5C 2 x 4 n -1
2n -+
+ ... + (2n + 1) C n x
1
Putting x = 1, we get
Here, q = 1 and r = 1
n × 2n - 1 × 2 + 2n = C 0 + 3C 1 + 5C 2 + ... + (2n + 1) C n Now, multiplying both sides by x in Eq. (i), then
or C 0 + 3C 1 + 5C 2 + ... + (2n + 1) C n = (n + 1) 2n nx (1 + x )n -1 = C 1 x + 2C 2 x 2 + 3C 3 x 3 + ... + n C n x n
I. Aliter Differentiating on both sides w.r.t. x, we get
LHS = C 0 + 3 C 1 + 5 C 2 + ... + (2n + 1) C n n { x × (n - 1) ( 1 + x )n - 2 + (1 + x )n - 1 × 1}
= C 0 + (1 + 2) C 1 + (1 + 4 ) C 2 + ... + (1 + 2n ) C n = C 1 × 1 + 2 2 C 2 x + 32 C 3 x 2
+ ... + n 2 C n x n -1
= (C 0 + C 1 + C 2 + ... + C n ) + 2 (C 1 + 2C 2 + ... + n C n ) Putting x =1, we get
= 2n + 2 × n × 2n - 1 = 2n + n × 2n [from Illusration 58] n {1 × (n - 1) × 2n - 2 + 2n -1 } = 12 × C 1 + 22 × C 2 + 32 × C 3
n
= (n + 1) 2 = RHS + ... + n 2 × C n
II. Aliter or 12 × C 1 + 22 × C 2 + 32 × C 3 + ... + n 2 × C n = n (n + 1) 2n - 2
LHS = C 0 + 3 C 1 + 5 C 2 + ... + (2n + 1) C n
n n n
Aliter
= S
r =0
(2r + 1) C r =n
S
r =0
n
2r × C r + S
r =0
n
Cr LHS = 12 × C 1 + 22 × C 2 + 32 × C 3 + ... + n 2 × C n
n n
n n
= S r 2 . n C r = S r 2 × × n - 1C r - 1
n
=2
r =0
S r × n Cr + S
r =0
n
Cr r =1 r =1 r
é n n n -1 ù
n
é
n
ù êëQ C r = r × Cr - 1ú
û
= 2 S r × × n -1C r -1 + S n C r êQ n C r = × n - 1C r
n n
- 1ú
r =0 r r =0 ë r û n n
n n =n S r × n - 1C r - 1 = n rS= 1 {(r - 1) + 1} × n -1C r - 1
= 2n S
r =0
n -1
Cr -1 +
r =0
S n
Cr
r =1

n n

= 2n (0 + n -1
C0 + n -1
C1 + n -1
C 2 + ... + n -1
Cn - 1) =n S (r - 1) × n - 1C r - 1 + n rS= 1 n -1C r - 1
r =1

+ ( C 0 + C 1 + C 2 + ... + n C n )
n n n n n

= 2n × 2n - 1 + 2n = (n + 1) × 2n = RHS
=n S (n - 1) × n - 2 C r - 2 + n rS= 1 n - 1C r - 1
r =1
 Chap 06 Binomial Theorem 461

n n = n (n - 1) { N C 0 + N C 1 + ... + N C N }
= n ( n - 1) S n - 2 C r - 2 + n rS= 1 n - 1C r -1
r =1 = n (n - 1) 2 N = n (n - 1) 2n - 2 = RHS
n-2 n-2 n-2
= n ( n - 1) ( 0 + C0 + C1 + C2 II. Aliter
+ ... + n-2
C n - 2 ) + n (n - 1 C 0 + n -1
C1 LHS = (1 × 2) C 2 + (2 × 3) C 3 + ... + {(n - 1) × n } C n
n

+ n -1
C 2 + ... + n -1
C n -1 ) = S
r =2
( r - 1) × r × n C r

= n (n - 1) × 2 n - 2 + n × 2 n - 1 = n (n + 1) 2n - 2 = RHS n
= S ( r - 1) × r × n × n - 1 × n - 2C r -2
n 2 r =2 r ( r - 1)
y Example 62. If (1 + x ) = C 0 + C 1 x + C 2 x
n
+ ... + C n x n , prove that (1 × 2) C 2 + (2 × 3) = ( n - 1) n S n - 2 C r -2
r =2
C 3 + K + {(n - 1) × n} C n = n (n - 1) 2n - 2 . n-2 n-2 n-2 n-2
= ( n - 1) n ( C0 + C1 + C 2 + ... + Cn - 2 )
Sol. Here, last term of n-2
= ( n - 1) n × 2 = RHS
(1 × 2) C 2 + (2 × 3) C 3 + ... + {(n - 1) × n } C n is (n - 1)n C n
i.e. (n - 1) n
y Example 63. If
[start with greater factor here greater factor is n] and last
(1 + x )n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + ... + C n x n , prove
term with positive sign, then n = n × 1 + 0
or n ) n (1 that C 0 - 2C 1 + 3C 2 - 4C 3 + ... + ( - 1)n (n + 1) C n = 0.
-n Sol. Numerical value of last term of
0 C 0 - 2C 1 + 3C 2 - 4C 3 + ... + ( - 1)n (n + 1) C n is
Here, q = 1 and r = 0 (n + 1) C n i.e., (n + 1), then
The given series is n + 1 = n × 1 + 1 or n ) n + 1 (1
(1 + x )n = C 0 + C 1 x + C 2 x 2
+ C3 x 3
+ ... + C n x n -n
Differentiating on both sides w.r.t. x , we get 1
n (1 + x )n -1 = 0 + C 1 + 2C 2 x + 3C 3 x 2 + ... + n C n x n -1 Here, q = 1 and r = 1
The given series is
Again, differentiating on both sides w.r.t. x, we get
(1 + x )n = C 0 + C 1x + C 2 x 2 + C 3 x 3
+... + C n x n
n (n - 1) (1 + x )n - 2 = 0 + 0 + (1 × 2) C 2 + (2 × 3) C 3 x
On multiplying both sides by x, we get
+ ... + {(n - 1) × n } C n x n - 2
x (1 + x )n = C 0 x + C 1x 2
+ C2 x 3
+ C3 x 4
+ ... + C n x n + 1
Putting x = 1, we get
On differentiating both sides w.r.t. x, we get
n (n - 1) (1 + 1)n - 2 = (1 × 2) C 2 + (2 × 3) C 3
x × n (1 + x )n -1 + (1 + x )n × 1 = C 0 + 2C 1x + 3C 2 x 2
+ ... + {(n - 1) n } × C n
+ 4 C 3 x 3 + ... + (n + 1) C n x n
or (1 × 2) C 2 + (2 × 3) C 3 + ... + {(n - 1) n } × C n = n (n - 1) 2n - 2
Putting x = - 1, we get
I. Aliter 0 = C 0 - 2C 1 + 3C 2 - 4C 3 + ... + ( - 1)n (n + 1) C n
LHS = (1 × 2)C 2 + (2 × 3)C 3 + (3 × 4)C 4
or C 0 - 2C 1 + 3C 2 - 4C 3 + ... + ( - 1)n (n + 1) C n = 0
+ ... + {(n - 1) n } × C n
n ( n - 1) n ( n - 1) ( n - 2) I. Aliter
= (1 × 2) + (2 × 3)
1 ×2 1 ×2 ×3 LHS = C 0 - 2C 1 + 3C 2 - 4C 3 + ... + ( -1)n (n + 1) C n
n ( n - 1) ( n - 2) ( n - 3) = C 0 - (C 1 + C 1 ) + (C 2 + 2C 2 ) - ( C 3 + 3C 3 )
+ (3 × 4 )
1 ×2 ×3 × 4 + ... + ( -1)n {C n + n C n }
+ ... + (n - 1) n × 1
= {C 0 - C 1 + C 2 - C 3 + ... + ( - 1)n C n }
ì ( n - 2) ( n - 2) ( n - 3) ü
= n (n - 1) í1 + + + ... + 1ý + { - C 1 + 2C 2 - 3C 3 + ... + ( - 1)n n C n }
î 1 1 × 2 þ
ì n ( n - 1) n (n - 1)(n - 2)ü
Now, in bracket, let n - 2 = N , then ï -n + 2 × -3 ï
= (1 - 1)n + í 1 ×2 1 ×2 ×3 ý
ì N N ( N - 1) ü ï + ... + ( -1)n × n ï
= n (n - 1) í1 + + + ... + 1ý î þ
î 1 2! þ
462 Textbook of Algebra

ì ( n - 1) ( n - 2) ü In bracket, put n - 1 = N , then

= 0 + n í - 1 + ( n - 1) - + ... + ( - 1)n ý
î 1 ×2 þ ì N N ( N - 1) ü
LHS = n í1 - + - ... + ( - 1) N ý
ì ( n - 1) ( n - 2) ü î 1 1 ×2 þ
= 0 - n í1 - (n - 1) + - ... + ( - 1)n -1ý
î 1 × 2 þ = n { N C 0 - N C 1 + N C 2 - ... + ( - 1)N N
CN }
Let in bracket, put n - 1 = N , we get N
= n (1 - 1) = 0 = RHS
ì N ( N - 1) ü
LHS = 0 - n í1 - N + - ... + ( - 1) N ý II. Aliter
î 1 × 2 þ
LHS = C 1 - 2C 2 + 3C 3 - ... + ( -1)n - 1 × n C n
N N N N N
= 0 - n { C 0 - C 1 + C 2 - ... + ( - 1) CN } n

= 0 - n ( 1 - 1) N
= 0 - 0 = 0 = RHS = S ( - 1)r - 1 × r × nC r
r =1
n
II. Aliter é n ù
LHS = C 0 - 2 C 1 + 3 C 2 - 4 C 3 + ... + ( - 1) (n + 1) C n n = S ( - 1)r - 1 × n × n - 1C r - 1 n n -1
êëQ C r = r × Cr - 1ú
r =1 û
n n n
= S ( -1)r (r + 1) n C r = r S= 0 ( - 1)r [r × n C r + n Cr ] =n S ( - 1)r - 1 × n - 1C r - 1
r =1
r =0
n
é n ù = n (1 - 1)n - 1 = 0 = RHS
= S ( -1)r [n ×n -1 C r -1 +n C r ]
r =0
n n -1
êëQ C r = r × C r -1 úû
y Example 65. If (1 + x ) n = C 0 + C 1 x + C 2 x 2
n n
=n S
r =0
( - 1)r × n - 1 C r -1 + S
r =0
( - 1)r × n C r + C 3 x 3 +... + C n x n
, prove that
n n C 0 - 3 C 1 + 5 C 2 - ... + ( -1)n (2n + 1) C n = 0.
= -n
r =0
S ( - 1) r -1
× n -1
Cr -1 + S
r =0
r
( - 1) × C r n
Sol. The numerical value of last term of
= - n (1 - 1)n - 1 + (1 - 1)n = 0 + 0 = 0 = RHS C 0 - 3C 1 + 5C 2 - ... + ( - 1)n (2n + 1) C n is (2n + 1) C n
i.e. (2n + 1)
y Example 64. If (1 + x )n = C 0 + C 1 x + C 2 x 2
and 2n + 1 = 2 × n + 1 or n ) 2n + 1 ( 2
+ C 3 x 3 +... + C n x n , prove that - 2n
C 1 - 2 C 2 + 3 C 3 - ... + ( - 1)n - 1 n C n = 0. 1
Here, q = 2 and r = 1
Sol. Numerical value of last term of
The given series is
C 1 - 2C 2 + 3C 3 - ... + ( - 1)n -1 n C n is nC n i.e., n, then 2 3
(1 + x )n = C 0 + C 1x + C 2 x + C3 x + ... + C n x n now,
and n = n × 1 + 0 or n ) n (1
replacing x by x 2 , then we get
-n
(1 + x 2 )n = C 0 + C 1x 2
+ C 2x 4
+ ... + C n x 2n
1
Here, q = 1 and r = 0 On multiplying both sides by x , we get
The given series is x (1 + x 2 )n = C 0 x + C 1x 3 + C 2 x 5 + ... + C n x 2n + 1
2 3
(1 + x )n = C 0 + C 1x + C 2 x + C3 x + ... + C n x n On differentiating both sides w.r.t. x , we get
On differentiating both sides w.r.t. x, we get x × n (1 + x 2 )n -1 2x + (1 + x 2 )n × 1 = C 0 + 3C 1x 2

n (1 + x )n -1 = 0 + C 1 + 2C 2 x + 3C 3 x 2
+ ... + n C n x n -1 + 5C 2 x 4 + ... + (2n + 1) C n x 2n

Putting x = - 1, we get Putting x = i in both sides, we get

n -1
0 = C 1 - 2C 2 + 3C 3 - ... + ( - 1) n Cn 0 + 0 = C 0 - 3C 1 + 5C 2 - ... + (2n + 1) ( -1)n C n
or C 1 - 2C 2 + 3C 3 - ... + ( - 1)n -1 n C n = 0 or C 0 - 3C 1 + 5C 2 - ... + ( - 1)n (2n + 1) C n = 0
I. Aliter I. Aliter
LHS = C 1 - 2C 2 + 3C 3 - ... + ( -1)n - 1 n × C n LHS = C 0 - 3C 1 + 5C 2 - ... + ( -1)n (2n + 1) C n
n ( n - 1) n ( n - 1) ( n - 2) = C 0 - (1 + 2) C 1 + (1 + 4 ) C 2 - ... + ( -1)n (1 + 2n ) C n
= n - 2× +3 - ... + ( - 1)n -1 × n × 1
1 ×2 1 ×2 ×3
= (C 0 - C 1 + C 2 - ... + ( - 1)n C n ) - 2 (C 1 - 2C 2
ì ( n - 1) ( n - 1) ( n - 2) ü
= n í1 - + - ... + ( - 1)n -1ý + ... + ( - 1)n - 1 n × C n )
î 1 1 × 2 þ
 Chap 06 Binomial Theorem 463

1
= (1 - 1)n - 2 × 0 [from Example 64] é C x 2
C2 x 3
Cn x n + 1 ù
Þ = êC 0 x + 1 + + ... + ú
= 0 = RHS ë 2 3 n +1 û0
II. Aliter
2 n +1 - 1 C C C
n
LHS = C 0 - 3C 1 + 5C 2 - ... + ( -1) (2n + 1) C n Þ = C 0 + 1 + 2 + ... + n
n +1 2 3 n +1
n n
= S ( - 1)r (2r + 1) n C r
r =1
= S ( - 1)r [2r × n C r
r =1
+ nC r ] or C 0 +
C1 C 2
+
C
+ ... + n =
2n + 1 - 1
2 3 n +1 n +1
n n
= 2 S n × n - 1C r -1 + S ( - 1)r × n C r I. Aliter
r =1 r =1
C1 C 2 C
= 2n (1 - 1)n - 1 + (1 - 1)n = 0 + 0 = 0 = RHS LHS = C 0 + + + ... + n
2 3 n +1
n n ( n - 1) 1
=1+ + + ... +
1 ×2 1 ×2 ×3 n +1
Use of Integration 1 é (n + 1)n (n + 1) n (n - 1) ù
= êë(n + 1) + 1 × 2 + + ... + 1ú
This method is applied only when the numericals occur as n +1 1 ×2 ×3 û
the denominator of the binomial coefficient. Put n + 1 = N , then
1 é N ( N - 1) N ( N - 1) ( N - 2) ù
LHS = êN + + + ... + 1ú
Solution Process N ë 2 ! 3 ! û
If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + ... + C n x n , =
1 N
[ C 1 + N C 2 + N C 3 + ... + N C N ]
then integrate both sides between the suitable limits N
which gives the required series. 1 2N - 1 2n + 1 - 1
= [(1 + 1)N - 1] = = = RHS
1. If the sum contains C 0 , C 1 , C 2 , . . . , C n are all positive N N n +1
signs, then integrate between limits 0 to 1. II. Aliter
n
2. If the sum contains alternate signs (i.e., +, –), then
+ ... + n = S
C1 C 2 C Cr
LHS = C 0 + +
integrate between limits –1 to 0. 2 3 n +1 r=0r +1
3. If the sum contains odd coefficients (i.e., n n n n +1
Cr + 1 é n + 1C r +1 n C r ù
C 0 , C 2 , C 4 , . . .), then integrate between –1 to +1. = S Cr
= S êQ = ú
r =0 ( r + 1) r = 0 ( n + 1) êë n +1 r + 1 úû
4. If the sum contains even coefficients (i.e.,
n
C 1 , C 3 , C 5 , . . .), then subtracting (2) from (1) and then =
1
S n +1
Cr +1
dividing by 2. ( n + 1) r = 0
5. If in denominator of binomial coefficient product of 1 n +1 n +1 n +1
= ( C1 + C2 + C3
two numericals, then integrate two times first times ( n + 1)
taken limits between 0 to x and second times take n +1
+ ... + Cn + 1)
suitable limits.
1 2n + 1 - 1
y Example 66. If (1 + x ) = C 0 + C 1 x n = (2n + 1 - 1) = = RHS
n +1 n +1
+ C 2 x 2+ ... + C n x n , prove that
y Example 67. If (1 + x )n = C 0 + C 1 x + C 2 x 2
C C C 2 n +1 - 1
C 0 + 1 + 2 + ... + n = . + C 3 x 3+ ... + C n x n , prove that
2 3 n+1 n+1 C C C 1
Sol. Q(1 + x )n = C 0 + C 1x + C 2 x 2 + ... + C n x n …(i)
C 0 - 1 + 2 - ... + ( - 1)n n = .
2 3 n+ 1 n+1
Integrating both sides of Eq. (i) within limits 0 to 1, then we get Sol. Q(1 + x )n = C 0 + C 1x + C 2 x 2 + ... + C n x n …(i)
1 1
ò0 ò0
n 2 n
(1 + x ) dx = (C 0 + C 1x + C 2 x + ... + C n x ) dx Integrating on both sides of Eq. (i) within limits - 1 to 0,
1 then we get
é (1 + x )n + 1 ù 0 0
ò-1(1 + x ) ò- 1(C 0 + C 1x + C 2 x
n 2
ê ú dx = + ... + C n x n ) dx
ë n + 1 û0
464 Textbook of Algebra

0 0 1
é (1 + x )n ù é C1 x 2
C2 x 3 C x n +1 ù = ( n + 1C 1 - n + 1C 2 + n + 1C 3 - ... + ( - 1)n × n + 1C n + 1 )
Þê ú = êC 0 x + + +...+ n ú ( n + 1)
ë n + 1 û -1 ë 2 3 n + 1 û- 1
=
1
{ n + 1C 0 - ( n + 1C 0 - n + 1C 1 + n + 1C 2 - n + 1C 3
( n + 1)
1-0 æ C C C ö + ... + ( - 1)n + 1 n + 1C n + 1 )}
Þ = 0 - ç - C 0 + 1 - 2 + ... + ( - 1)n +1 n ÷
n +1 è 2 3 n + 1ø 1 1 1
= [ 1 - (1 - 1)n + 1 ] = [1 - 0] = = RHS
1 C C Cn ( n + 1) ( n + 1) n +1
Þ = C 0 - 1 + 2 - ... + ( - 1)n + 2
n +1 2 3 n +1
y Example 68. If (1 + x )n = C 0 + C 1 x + C 2 x 2
1 C C Cn
Þ = C 0 - 1 + 2 - ... + ( - 1)n + C 3 x 3 +... + C n x n , prove that
n +1 2 3 n +1
[Q ( - 1)n + 2 = ( - 1)n ( - 1)2 = ( - 1)n ] C0 C2 C4 2n
+ + + ... = .
C1 C 2 C 1 1 3 5 n+1
Hence, C 0 - + - ... + ( - 1)n n = 2 3
2 3 n +1 n + 1 Sol. Q (1 + x )n = C 0 + C 1x + C 2 x + C3 x
I. Aliter + C 4 x 4 +... + C n x n …(i)
C1 C 2 C
LHS = C 0 - + - ... + ( - 1)n n Integrating on both sides of Eq. (i) within limits - 1 to 1,
2 3 n +1 then we get
n n ( n - 1) 1 1 1 1
=1- + - ... + ( - 1)n = ò- 1 (1 + x ) ò-1(C 0 + C 1x + C 2 x
n 2 3 4
dx = + C3 x + C4 x
2 1 ×2 ×3 n + 1 ( n + 1)
( n + 1) n ( n + 1) n ( n - 1) + ... + C n x n ) dx
é ù
êë (n + 1) - 1 × 2 + - ... + ( - 1)n ú 1 1
1 ×2 ×3 û ò-1(C 0 + C 2 x + C 4 x 4 + ...)dx + ò (C 1x + C 3 x
2 3
= + ...)dx
-1
Put n + 1 = N , we get 1
= 2 ò (C 0 + C 2 x 2
+ C 4 x 4 + ...) dx + 0
é N ( N - 1) N ( N - 1) ( N - 2) ù 0
1 N - +
= ê 1 ×2 1 ×2 ×3 ú [by property of definite integral]
Nê N -1 ú
[since, second integral contains odd function]
ë - ... + ( - 1) û
1 1
1 N -1
é (1 + x )n + 1 ù éæ C2 x 3
C4 x 5 öù
= [ C 1 - N C 2 + N C 3 - ... + ( - 1)N ] ê ú = 2 ê çC 0 x + + + ...÷ ú
N ë n + 1 û -1 ëê è 3 5 ø úû 0
1 N
2n + 1
= - [ - N C 1 + N C 2 - N C 3 + ... + ( - 1) N C N ] æ C C ö
N Þ = 2 çC 0 + 2 + 4 + ...÷
n +1 è 3 5 ø
1 N C2 C4 2n
= - [ C 0 - N C1 + N C 2 - N C 3 or C 0 + + +K=
N 3 5 n +1
N
+ ... + ( - 1)N C N - NC 0 ] I. Aliter
1 1 1 C2 C4
=- [(1 - 1)N _ N C 0 ] = - [0 - 1] = LHS = C 0 + + + ...
N N N 3 5
n ( n - 1) n ( n - 1) ( n - 2) ( n - 3)
=
1
= RHS =1+ + + ...
n +1 1 ×2 ×3 1 ×2 ×3 × 4 ×5
1 ì n + 1 (n + 1)n (n - 1)
II. Aliter = í +
n
( n + 1) î 1 1 ×2 ×3
( - 1)r × C r
- ... + ( - 1)n n = S
C1 C 2 C
LHS = C 0 - + (n + 1)n (n - 1) (n - 2) (n - 3) ü
2 3 n +1 r = 0 r + 1 + + ...ý
n
1 ×2 ×3 × 4 ×5 þ
r
× nC r
= S ( - 1) =
1
{ n + 1C1 + n +1
C3 + n +1
C 5 + ...}
r =0 r +1
n +1
n +1 é n +1
Cr ù
n n
Cr C r +1
= S ( - 1)r ×
+1
êQ = ú =
1
[sum of even binomial coefficients of (1 + x )n + 1]
r =0 ( n + 1) êë n + 1 r + 1 úû
( n + 1)
n
=
1
S ( - 1)r × n + 1C r +1 =
2n + 1 - 1
=
2n
= RHS
( n + 1) r = 0 n +1 n +1
 Chap 06 Binomial Theorem 465

C0 C2 C4 Put n + 1 = N , then
II. Aliter LHS = + + + ...
1 3 5 1 é N ( N - 1) N ( N - 1) ( N - 2) ( N - 3)
LHS = +
Case I If n is odd say n = 2m + 1, " m ÎW , then N êë 2! 4!
m 2 m +1 m 2m + 2
C 2r + 1 N ( N - 1)( N - 2) ( N - 3) ( N - 4 ) ( N - 5) ù
LHS = S C 2r
= S + + ...ú
r =0 2r + 1 r =0 (2m + 1) 6! û
é 2 m +1C 2r 2m+ 2
+1ù
C 2r 1 N
êQ = ú = [ C 2 + N C 4 + N C 6 + ... ]
2r + 1 2m +1 N
êë úû
1
1 2n = [ ( N C 0 + N C 2 + N C 4 + N C 6 + ...) - N C 0 ]
= ×22 m + 2 -1 = = RHS N
(2m + 1) n +1
1 -1 2n - 1
[Qn = 2m + 1] = [2 N - 1] = = RHS
N n +1
Case II If n is even say n = 2m, " m Î N , then
II. Aliter
m 2m m 2 m +1
C 2r C 2r
LHS = S = S +1
LHS =
C1 C 3 C 5
+ + + ...
r =0 2r + 1 r =0 (2m + 1) 2 4 6
é 2 m +1
C 2r 2m
C 2r ù Case I If n is odd say n = 2m + 1 , " m ÎW , then
+1
êQ = ú m 2 m +1 m 2m+2
2m + 1 2r + 1 úû C2r C 2r
êë LHS = S +1
= S +2
r =0 2r + 2 r =0 (2m + 2)
22 m + 1 - 1 2n
= = = RHS [Qn = 2m ] é 2m+2
C 2r +2
2 m +1
C 2r +1ù
2m + 1 n +1 êQ = ú
êë 2m + 2 2r + 2 úû
y Example 69. If (1 + x )n = C 0 + C 1 x + C 2 x 2+ C 3 x 3
1 2 m +2 2 m +2 2 m +2
= ( C2 + C 4 + ... + C2 m + 2)
C C C 2n - 1 ( 2m + 2)
+... + C n x n , prove that 1 + 3 + 5 + ... = .
2 4 6 n+1 1 2n - 1
= × ( 22 m + 2 - 1 - 2 m +1
C0) = [Q2m + 1 = n ]
Sol. We know that, from Examples (66) and (67) ( 2m + 2) n +1
C1 C 2 C 3 C 4 C 5 2n + 1 - 1 = RHS
C0 + + + + + + ... = …(i)
2 3 4 5 6 n +1 Case II If n is even say n = 2m, " m Î N , then
m -1 2m m -1 2 m +1
C1 C 2 C 3 C 4 C 5 1 C 2r C 2r
and C 0 -
2
+
3
-
4
+
5
-
6
+ ... =
n +1
…(ii) LHS = S +1
= S +2
r =0 (2r + 2) r =0 (2m + 1)
On subtracting Eq. (ii) from Eq. (i), we get é 2 m +1
+1ù
2m
C 2r +2 C 2r
n +1 êQ = ú
æC C C ö 2 -2 2m + 1 2r + 2
2 ç 1 + 3 + 5 + ...÷ = êë úû
è2 4 6 ø n +1 m -1
On dividing each sides by 2, we get =
1
S 2 m + 1C 2 r + 2
(2m + 1) r = 0
C1 C 3 C 5 2n - 1 1
+ + + ... = = ( 2 m +1C 2 + 2 m +1C 4 + 2 m +1
C6
2 4 6 n +1 (2m + 1)
2 m +1
C1 C 3 C 5 + ... + C2 n )
I. Aliter LHS = + + + ...
2 4 6 1
= × ( 22 m + 1 - 1 - 2m +1
C0)
n n ( n - 1) ( n - 2) ( 2m + 1)
= +
1 ×2 1 ×2 ×3 × 4 2n - 1
= = RHS [Qn = 2m ]
n ( n - 1) ( n - 2) ( n - 3) ( n - 4 ) n +1
+ + ...
1 ×2 ×3 × 4 ×5 ×6
1 é ( n + 1) n ( n + 1) n ( n - 1) ( n - 2) y Example 70. If (1 + x )n = C 0 + C 1 x
= +
n + 1 êë 1 × 2 1 ×2 ×3 × 4 + C 2 x 2+ ... + C n x n , prove that
n +1
( n + 1) n ( n - 1) ( n - 2) ( n - 3) ( n - 4 ) ù C 1 33 C 2 3 4 C 3 3n +1 C n 4 - 1
+ + ...ú 3C 0 + 3 2 + + + ... + = .
1 ×2 ×3 × 4 ×5 ×6 û 2 3 4 n+1 n+1
466 Textbook of Algebra

2 3 n
Sol. Q (1 + x )n = C 0 + C 1x + C 2 x + C3 x +... + C n x n …(i)
=
1
S n +1
Cr +1 × 3r +1

Integrating on both sides of Eq. (i) within limits 0 to 3, we get ( n + 1) r = 0

3 3 1 n +1 n +1 n +1
ò0 (1 + x ) dx = ò0 ( C 0 + C 1x + C 2 x = C1 × 3 + C 2 × 32 + C 3 × 33
2
n
+ C 3 x 3 +...+ C n x n )dx (
( n + 1)
3 n +1
é (1 + x )n + 1 ù é C1 x 2
C2 x 3
C3 x 4
+ ... + C n + 1 × 3n +1 )
Þê ú = êC 0 x + + + +
ë n + 1 û0 ë 2 3 4 1
= [ (1 + 3)n + 1 - n +1
C0]
n +1ù
3 ( n + 1)
C x
K+ n ú
n + 1 û0 4n + 1 - 1
= = RHS
n +1
4n + 1 - 1 32 C 1 33 C 2 34 C 3 3n + 1 C n
Þ = 3C 0 + + + + ... +
n +1 2 3 4 n +1 y Example 71. If (1 + x )n = C 0 + C 1 x + C 2 x 2+... + C n x n ,
Hence, 22 23 24 2n + 2 C n
32 C 1 33 C 2 34 C 3 3n + 1 C n 4n + 1 - 1 show that C0 + C1 + C 2 + ... +
3C 0 + + + + ... + = 1× 2 2× 3 3×4 (n + 1) (n + 2)
2 3 4 n +1 n +1 n +2
3 - 2n - 5
I. Aliter = .
(n + 1) (n + 2)
32 C 1 33 C 2 34 C 3 3n + 1 C n
LHS = 3C 0 + + + + ... + Sol. Given,
2 3 4 n +1
2 3
(1 + x )n = C 0 + C 1 x + C 2 x + C3 x +... + C n x n
32 × n 33 × n (n - 1) 34 × n (n - 1) (n - 2)
= 3 ×1 + + + ...(i)
2 1 ×2 ×3 1 ×2 ×3 × 4
3n +1 Integrating both sides of Eq. (i) within limits 0 to x , we get
+ ... +
n +1 x x
ò0 (1 + x ) ò0 ( C 0 + C 1x + C 2 x
n 2
dx = + ... + C n x n ) dx
1 é 2 3
3 ( n + 1) n 3 ( n + 1) n ( n - 1)
= ê3 × ( n + 1) + + é (1 + x )n + 1 ù
x
( n + 1) ë 1 ×2 1 ×2 ×3 Þ ê ú
34 ( n + 1) n ( n - 1) ( n - 2) ù ë n + 1 û0
+ + ... + 3n + 1 ú x
1 ×2 ×3 × 4 û é Cx 2
C x 3
C x n +1ù
= êC 0 x + 1 + 2 +K+ n ú
Put n + 1 = N , then ë 2 3 n + 1 û0
1 é 32 N ( N - 1) 33 N ( N - 1) ( N - 2)
LHS = ê3N + + (1 + x )n + 1 - 1 Cx 2
C2 x 3
Cn x n + 1
N ë 2! 3! Þ = C 0x + 1 + + ... +
( n + 1) 2 3 n +1
34 N ( N - 1) ( N - 2) ( N - 3) ù …(ii)
+ + ... + 3N ú
4! û Again, integrating both sides of Eq. (ii) within limits 0 to 2,
1 N we get
= [ C 1 (3) + N C 2 (3)2 + N C 3 (3)3 + ... + N C N (3)N ]
N 2( 1 + x )n + 1 - 1
=
1 N
[ C 0 + NC 1 ( 3) + N C 2 ( 3) 2 + N C 3 ( 3) 3
ò0 ( n + 1)
dx
N
+ ... + N C N (3) N - N C 0 ] 2æ C 1x 2
C2 x 3
Cn x n + 1 ö
= ò0 çC 0 x +
è
+ + ... +
n +1 ø
÷ dx
1 4 N - 1 4n + 1 - 1 2 3
= { ( 1 + 3) N - 1 } = = = RHS
N N n +1 2
1 æ (1 + x )n + 2 öù é C x 2 C 1x 3 C 2 x 4
II. Aliter Þ ç - x ÷ú = ê 0 + +
( n + 1) è n + 2 ø ûú 0 ë 1 × 2 2 ×3 3× 4
C 1 33 C 2 34 C 3 3n + 1 C n
LHS = 3C 0 + 32 + + + ... + 2
2 3 4 n +1 Cn x n + 2 ù
+K+ ú
n r +1 n +1 n +1 ( n + 1) ( n + 2) û 0
× nC r 3r × Cr
= S3 = S
+1
r =0 ( r + 1) r =0 ( n + 1) 1 ì 3n + 2 1 ü 22 23 24
Þ í -2- ý= C0 + C1 + C2
é n +1
Cr +1 Cr ù
n ( n + 1) î n + 2 n + 2þ 1 × 2 2 ×3 3× 4
êQ = ú
êë n +1 r + 1 úû 2n + 2 C n
+ ... +
( n + 1) ( n + 2)
 Chap 06 Binomial Theorem 467

22 23 24 2n + 2 C n
Hence, C0 + C1 + C 2 + ... + When Each Term in Summation Contains
1 ×2 2 ×3 3× 4 ( n + 1) ( n + 2)
3 n+2
- 2n - 5
the Product of Two Binomial Coefficients
=
( n + 1) ( n + 2) or Square of Binomial Coefficients
I. Aliter Solution Process
22 23 24 2n + 2 C n
LHS = C0 + C1 + C2 + K + 1. If difference of the lower suffixes of binomial
1 ×2 2 ×3 3× 4 ( n + 1) ( n + 2) coefficients in each term is same.
22 23 24 n ( n - 1) 2n + 2 × 1 i.e. n
C 0 n C 2 + n C 1 × n C 3 + n C 2 × n C 4 + ...
= ( 1) + ×n + + ... +
1 ×2 2 ×3 3× 4 1 ×2 (n + 1) (n + 2)
Here, 2 - 0 = 3 - 1 = 4 - 2 = .... = 2
1 ì ( n + 2) ( n + 1) 2 ( n + 2) ( n + 1) n 3
= í 2 + 2 Case I If each term of series is positive, then
( n + 1) ( n + 2) î 1 ×2 1 ×2 ×3
(1 + x ) n = C 0 + C 1 x + C 2 x 2 + ... + C n x n …(i)
( n + 2) ( n + 1) n ( n - 1) 4 ü
+ 2 + ... + 2n + 2 ý
1 ×2 ×3 × 4 þ Interchanging 1 and x , we get
Put n + 2 = N , then we get ( x + 1) n = C 0 x n + C 1 x n - 1 + C 2 x n -2
+... + C n …(ii)
1 ì N ( N - 1) 2 N ( N - 1) ( N - 2) 3 Then, multiplying Eqs. (i) and (ii) and equate the
= í 2 + 2
N ( N - 1) î 1 × 2 1 ×2 ×3 coefficients of suitable power of x on both sides.
N ( N - 1) ( N - 2) ( N - 3) 4 ü 1
+ 2 + ... + 2N ý Replacing x by in Eq. (i), then we get
1 ×2 ×3 × 4 þ x
1 n
= { N C 2 ( 2) 2 + N C 3 ( 2) 3 + N C 4 ( 2) 4 æ 1ö C1 C2 C
N ( N - 1) ç1 + ÷ = C 0 + + + ... + n …(iii)
+ ... + N C N (2)N ] è xø x x 2
xn
1 Then, multiplying Eqs. (i) and (iii) and equate the
= { N C 0 + N C 1(2) + N C 2 (2)2 + N C 3 (2)3
N ( N - 1) coefficients of suitable power of x on both sides.
+ N C 4 (2)4 + ... + N C N (2)N - N C 0 - N C 1 (2)]
y Example 72. If (1 + x )n = C 0 + C 1 x + C 2 x 2
1
= {(1 + 2)N - 1 - 2N } + C 3 x 3 +... + C n x n , prove that
N ( N - 1)
3n + 2 - 1 - 2 (n + 2) 3n + 2 - 2n - 5 C 0C r + C 1C r +1 + C 2 C r + 2 + ... + C n - r C n
= = = RHS 2n !
( n + 2) ( n + 1) ( n + 1) ( n + 2) = .
(n - r )! (n + r )!
II. Aliter
22 23 24 2 n + 2× C n Sol. Here, differences of lower suffixes of binomial coefficients
LHS = ×C 0 + ×C1 + × C 2 + ... + in each term is r .
1 ×2 2 ×3 3× 4 ( n + 1) ( n + 2)
n +1
i.e., r - 0 = r + 1 - 1 = r + 2 - 2 = .... = n - (n - r ) = r
2r + 1 n
= S × Cr -1
Given,
r =1 r ( r + 1) (1 + x )n = C 0 + C 1x + C 2 x 2 + ... + C n - r x n - r + ... + C n x n
n +1 r +1
× n + 2 Cr é n+2 n
Cr - 1 ù
= S2 +1
êQ
Cr + 1
= ú Now,
…(i)
r =1 ( n + 1) ( n + 2) êë ( n + 1 ) ( n + 2 ) r ( r + 1) úû
n +1 ( x + 1)n = C 0 x n + C 1 x n - 1 + C 2 x n - 2 + ... + C r x n - r
=
1
S n + 2 C r + 1 × 2r + 1 + Cr +1x
n - r -1
+ Cr x n -r -2
+ ... + C n …(ii)
( n + 1) ( n + 2) r = 1 +2

1 On multiplying Eqs. (i) and (ii), we get

= [ n + 2 C 2 × 22 + n + 2 C 3 × 23
( n + 1) ( n + 2) (1 + x )2n = (C 0 + C 1x + C 2 x 2
+ ... + C n - r x n - r + ...
+ ... + n + 2 C n + 2 × 2n + 2 ]
1 + C n x n ) ´ (C 0 x n + C 1 x n - 1
= [ (1 + 2)n + 2 - n+2
C0 - n+2
C 1 × 21 ]
( n + 1) ( n + 2) + C 2 x n - 2 + ... + C r x n - r + C r +1x
n - r -1

(3 n + 2 - 2n - 5) + Cr +2 x n -r -2
+ ... + C n ) …(iii)
= = RHS
( n + 1) ( n + 2)
468 Textbook of Algebra

Now, coefficient of x n - r on LHS of Eq. (iii) = 2n

C n -r Corollary II For r = 1,
2n !
=
2n ! C 0C 1 + C 1C 2 + C 2 C 3 + ... + C n -1 C n =
(n - r ) ! (n + r ) ! ( n - 1) ! ( n + 1) !
Corollary III For r =2,
and coefficient of x n - r on RHS of Eq. (iii)
2n !
= C 0C r + C 1C r + 1 + C 2 C r + 2 + ... + C n - r C n C 0C 2 + C 1C 3 + C 2 C 4 + ... + C n - 2 C n =
( n - 2) ! ( n + 2) !
But Eq. (iii) is an identity, therefore coefficient of x n - r in
RHS = coefficient of x n - r in LHS. y Example 73. If (1 + x )n = C 0 + C 1 x
Þ C 0C r + C 1C r +1 + C 2 C r + 2 + ... + C n - r C n + C 2 x 2 +... + C n x n , prove that
=
2n ! 2n ! 1 × 3 × 5 ...(2n - 1) n
(n - r )! (n + r ) ! C 02 + C12 + C 22 = = ×2 .
n!n! n!
Aliter
Sol. Given, (1 + x )n = C 0 + C 1x + C 2 x 2 +... + C n x n …(i)
Given,
n n n -1 n-2
(1 + x )n = C 0 + C 1 x + C 2 x 2
+ ... + C r x r + C r r +1 Now, ( x + 1) = C 0 x + C 1x + C 2x + ... + C n …(ii)
+1x
+2 On multiplying Eqs. (i) and (ii), we get
+ Cr +2 xr + ... + C n - r x r + ... + C n x n …(i)
(1 + x )2n = ( C 0 + C 1x + C 2 x 2 +... + C n x n )
n
æ 1ö C C C Cr + 1 Cr + 2
Now, ç1 + ÷ = C 0 + 1 + 22 + ... + rr + r + 1 + r + 2 ´ (C 0 x n + C 1 x n - 1 + C 2 x n - 2 + ... + C n ) …(iii)
è x ø x x x x x
Cn - r Cn Now, coefficient of x n in RHS
+ ... + + ... + …(ii) 2 2 2 2
xn - r xn = C 0 + C 1 + C 2 + ... + C n
On multiplying Eqs. (i) and (ii), we get 2n 2n !
And coefficient of x n in LHS = Cn =
(1 + x ) 2n n !n !
2 r +1
= (C 0 + C 1 x + C 2 x + ... + C r x r + C r +1x
xn 1 × 2 × 3 × 4 × 5...× (2n - 1) 2n 1 × 3 × 5... (2n - 1) 2n n !
= =
+2
+ Cr +2 xr + .. + C n - r x n - r + .. + C n x n ) n !n ! n !n !

æ C C C Cr + 1 Cr + 2 But Eq. (iii) is an identity, therefore coefficient of x n in RHS

´ çC 0 + 1 + 22 + ... + rr + r + 1 + r + 2 = coefficient of x n in LHS.
è x x x x x
Cn ö 2 2 2 2 2n !
+ ... +
Cn - r
+ ... + Þ C 0 + C 1 + C 2 + ... + C n =
÷ …(iii) n !n !
xn - r xn ø
1 × 3 × 5 ×...× (2n - 1) n
1 = ×2
Now, coefficient of in RHS n!
xr Aliter
= (C 0C r + C 1C r + 1 + C 2 C r + 2 + ... + C n - r C n ) Given, (1 + x )n = C 0 + C 1 x + C 2 x 2
+ ... + C n x n …(i)
1
\ Coefficient of r in LHS = Coefficient of x n - r in æ 1ö C C
n
C
x Now, ç1 + ÷ = C 0 + 1 + 22 + ... + nn …(ii)
2n 2n 2n ! è xø x x x
(1 + x ) = C n - r =
( n - r ) ! (n + r ) ! On multiplying Eqs. (i) and (ii), we get
1 (1 + x )2n
But Eq. (iii) is an identity, therefore coefficient of in n
= ( C 0 + C 1x + C 2 x 2 + ... + C n x n )
xr x
æ C C C ö
1 ´ çC 0 + 1 + 22 + ... + nn ÷ …(iii)
RHS = coefficient of in LHS. è x x x ø
xr
Now, constant term in RHS = C 02 + C 12 + C 22 + ... + C n2
Þ C 0C r + C 1C r + C 2 C r + 2 + ... + C n -r C n
+1
2n ! (1 + x )2n
= Constant term in LHS = Constant term in
(n - r )! (n + r ) ! xn
2n !
Corollary I For r = 0, = Coefficient of x n in (1 + x )2n = 2n
Cn =
n !n !
2 2 2 2 2n !
C 0 + C 1 + C 2 + ... + C n = n ! 2n [1 × 3 × 5... (2n -1)] 2n [1 × 3 × 5... (2n - 1)]
(n !)2 = =
n !n ! n!
 Chap 06 Binomial Theorem 469

2n 2n 2n
But Eq. (iii) is an identity, therefore the constant term in æ 1ö 2n C1 2n C 2 C 2n
RHS = constant term in LHS. and ç1 - ÷ = C0 - + - ... + …(ii)
è xø x x2 x 2n
2 2 2 2 2n ! {1 × 3 × 5... (2n - 1)} n
Þ C 0 + C 1 + C 2 + ... + C n = = 2 On multiplying Eqs. (i) and (ii), we get
n !n ! n!
( x 2 - 1) 2 n 2n 2n 2n
Case II If terms of the series alternately positive and 2n
=( C0 + C 1x + C 2 x 2 + ... + 2n
C 2n x 2n
)
x
negative, then 2n 2n 2n
2n C1 C2 C 2n
n
(1 - x ) = C 0 - C 1 x + C 2 x 2
- ... + ( -1) C n x n n
…(i) ´( C0 - + 2
- ... + ) …(iii)
x x x 2n
n -1 n -2
and ( x + 1) n = C 0 x n + C 1 x + C2 x + ... + C n …(ii) Now, constant term in RHS
Then, multiplying Eqs. (i) and (ii) and equate the = ( 2 n C 0 )2 - ( 2 n C 1 )2 + ( 2 n C 2 )2 - ... + ( 2 n C 2n )2
coefficient of suitable power of x on both sides. ( x 2 - 1) 2 n
Constant term in LHS = Constant term in
Or x 2n
1
Replacing x by in Eq. (i), we get = Coefficient of x 2n
in ( x - 1)2 2n
x
n = Coefficient of x 2 n in (1 - x 2 )2 n
æ 1ö C1 C2 C
ç1 - ÷ = C 0 - + - ... + ( - 1) n n ...(iii) 2n 2n
è x ø x x 2
xn = C n ( - 1)n = ( - 1)n × Cn

Then, multiplying Eqs. (i) and (iii) and equate the But Eq. (iii) is an identity, therefore the constant term in
RHS = constant term in LHS.
coefficient of suitable power of x on both sides. 2n
Þ( C 0 )2 - ( 2n
C 1 )2 + ( 2n
C 2 )2 - ... + ( 2n
C 2 n )2
y Example 74. Prove that = ( - 1)n × 2n
Cn
( 2n C 0 ) 2 - ( 2n C1 ) 2 +( 2n C 2 ) 2 - ... + ( 2n C 2n ) 2 = (- 1)n × 2n
Cn .
y Example 75. If (1 + x )n = C 0 + C 1 x
Sol. Since, (1 - x )2 n = 2n
C0 - 2n
C1 x + 2n
C2 x 2

+ C 2 x 2 +... + C n x n , prove that

- ... + ( - 1)2 n × 2n
C 2 n x2 n
C 0 2 - C 1 2 + C 2 2 - ... + ( -1)n × C n2 = 0 or
or (1 - x )2 n = 2n
C0 - 2n
C1 x + 2n
C 2 x 2 - ... +
C 2n x 2n 2n
n!
…(i) ( - 1)n / 2 × , according as n is odd or even.
and ( x + 1)2 n = 2n
C 0 x 2n + 2 nC 1 x 2 n - 1 + 2 nC 2 x 2 n - 2 (n / 2)! (n / 2)!
+ ... + 2 n C 2 n …(ii) Also, evaluate C 0 2 + C 1 2 + C 2 2 - ... + ( - 1)n × C n 2 for n
On multiplying Eqs. (i) and (ii), we get = 10 and n = 11
( x 2 - 1)2 n = ( 2 n C 0 - 2 n C 1 x + 2 n C 2 x 2 - ... + 2 n C 2n x 2 n ) Sol. Since, (1 - x )n = C 0 - C 1x + C 2 x 2 -... + ( - 1)n C n x n …(i)
´ ( 2 n C 0 x 2 n + 2 n C 1 x 2 n - 1 + 2 n C 2 x 2 n - 2 + ... + 2 n C 2 n ) and ( x + 1)n = C 0 x n + C 1 x n - 1 + C 2 x n - 2 +... + C n …(ii)

Now, coefficient of x 2 n in RHS …(iii) On multiplying Eqs. (i) and (ii), we get
2n 2 2n 2 2n
= ( C 0 ) - ( C 1 ) + ( C 2 ) - ... + ( C 2 n ) 2 2n 2 (1 - x 2 )n = {C 0 - C 1x + C 2 x 2
- ... + ( -1)n C n x n }
´ ( C 0 x n + C 1 x n - 1 + C 2 x n - 2 +... + C n ) …(iii)
Now, LHS can also be written as (1 - x 2 )2 n .
\ General term in LHS, Tr = 2n
C r ( - x 2 )r Now, coefficient of x n in RHS
+1
2n = C 02 - C 12 + C 22 - ... + ( - 1)n C n2
Putting r = n , we get Tn + 1 = ( - 1)n × Cn x 2n
General term in LHS = Tr +1 =
n
C r ( - x 2 ) r = n C r ( -1)r x 2r
Þ Coefficient of x 2 n in LHS = ( - 1)n × 2n
Cn
Putting 2r = n , we get r = n / 2
But Eq. (iii) is an identity, therefore coefficient of x 2 n in
\ T (n / 2 ) + 1 = n C n / 2 ( - 1)n / 2 x n
RHS = coefficient of x 2 n in LHS
\ Coefficient of x n in LHS = n C n / 2 ( - 1)n / 2
2n 2 2n 2 2n 2 2n 2
Þ ( C 0 ) - ( C 1 ) + ( C 2 ) - ... + ( C 2 n ) n!
= ( - 1)n / 2 ×
2n
= ( - 1)n × Cn ( n / 2) ! ( n / 2) !
Aliter ì0, if n is odd
ï éQ æ odd ö ! = ¥ ù
Since, (1 + x )2 n = =í ê çè 2 ÷ø
2n 2n 2n
C0 + C 1x + C 2x 2 n ! ú
( - 1)n / 2 , if n is even ë û
+ ... + 2n
C 2 n x 2 n …(i) ï ( n / 2) ! ( n / 2) !
î
470 Textbook of Algebra

But Eq. (iii) is an identity, therefore coefficient of x n in RHS 2. If sum of the lower suffixes of binomial
= coefficient of x n in LHS. coefficients in each term is same.
Þ C 02 - C 12 + C 22 - ... + ( - 1)n C n2 i.e., C 0 C n + C 1 C n - 1 + C 2 C n - 2 + ... + C n C 0
ì0 , if n is odd Here, 0 + n = 1 + (n - 1) = 2 + (n - 2 ) = ... = n + 0 = n
ï
=í n! Case I If each term of series is positive, then
( - 1)n / 2 , if n is even
ï ( / 2 ) ! ( n / 2) ! 2
î n (1 + x ) n = C 0 + C 1 x + C 2 x +... + C n x n
…(i)
Now, for n = 10, 2
10 !
and (1 + x ) n = C 0 + C 1 x + C 2 x +... + C n x n
…(ii)
C 02 - C 12 + C 22 - ... + C 10
2
= ( - 1)10 / 2 = - 252
5!5! Then, multiplying Eqs. (i) and (ii) and equate the
[Q10 is even] coefficient of suitable power of x on both sides.
and from n = 11, y Example 76. Prove that
m +n
C 02 - C 12 + C 22 - ... - C 11
2
=0 [Q11 is odd] Cr = m Cr + m Cr -1 n C1 + m
C r -2 n C 2 + ... + n C r
Aliter if r < m , r < n and m, n, r are positive integers.
n 2 n
Since, (1 + x ) = C 0 + C 1x + C 2 x +... + C n x …(i) Sol. Here, sum of lower suffixes of binomial coefficients in
each term is r .
1
Replacing x by - , then we get i.e. r = r - 1 + 1 = r - 2 + 2 = ... = r = r
x
Since,
n
æ 1ö C1 C 2 C
+ 2 - ... + ( - 1)n nn (1 + x )m = mC 0 + mC 1 x +... + mC r - 2 x r - 2 + mC r - 1 x r -1
ç1 - ÷ = C 0 - …(ii)
è x ø x x x m r m
+ C r x + ... + C m x …(i) m

On multiplying Eqs. (i) and (ii), we get and (1 + x ) = C 0 + C 1 x + C 2 x 2 + ... + n C r x r

n n n n
2 n
( x - 1) 2 + ... + n C n x n ...(ii)
n
= (C 0 + C 1 x + C 2 x + ... + C n x n ) ´
x On multiplying Eqs. (i) and (ii), we get
æ C1 C 2 n Cn ö
(1 + x )m + n = ( mC 0 + mC 1x + ... + mC r - 2 x r - 2 + mC r - 1x r - 1
çC 0 - + 2 - ... + ( - 1) n ÷ …(iii)
è x x x ø + mC r x r + ... + mC m x m ) ´ (n C 0 + n C 1x + n C 2 x 2
Now, constant term in RHS + ... + n C r x r + ... + n C n x n ) …(iii)
r
= C 10 - C 12 + C 22 - ... + ( -1)n C n2 Now, coefficient of x in RHS
= mC r × n C 0 +m C r - 1 × n C 1 +m C r - 2 × n C 2 + ... + mC 0 × n C r
\Constant term in LHS
= mC r + mC r -1 × n C 1 +m C r -2 × n C 2 + ... + n C r
( x 2 - 1)n
= Constant term in Coefficient of x r in LHS = m +n
Cr
xn
2
But Eq. (iii) is an identity, therefore coefficient of x r in LHS
= Coefficient of x n in ( x - 1)n = coefficient of x r in RHS.
= Coefficient of x n in n C n / 2 ( x 2 )n - (n / 2 ) ( - 1)n / 2 Þ m +n
C r = mC r + mC r -1×
n
C 1 + mC r -2 × n C 2 + ... + n C r
= ( - 1)n / 2 × n C n / 2 Case II If terms of the series alternately positive and
n! negative, then
= ( - 1)n / 2 ×
( n / 2) ! ( n / 2) ! (1 - x ) n = C 0 - C 1 x + C 2 x 2 - ... + ( - 1) n C n x n …(i)
ì 0, if n is odd
and (1 + x ) n = C 0 + C 1 x + C 2 x 2 + ... + C n x n …(ii)
ï
=í n!
( -1)n / 2 × , if n is even Then, multiplying Eqs. (i) and (ii) and equate the
ï ( n / 2 ) ! ( n / 2) !
î
coefficient of suitable power of x on both sides.
But Eq. (iii) is an identity, therefore the constant term in
RHS = constant term in LHS. y Example 77. If (1 + x )n = C 0 + C 1 x
Þ C 02 - C 12 + C 22 - ... + ( -1)n C n2 + C 2 x 2 +... + C n x n , prove that
ì0, if n is odd C 0 C n - C 1 C n - 1 + C 2 C n - 2 - ... + ( - 1)n C n C 0 = 0 or
ï
=í n ! n!
( -1)n / 2 × , if n is even ( - 1)n / 2 , according as n is odd or even.
ï ( n / 2) ! ( n / 2) !
î (n / 2)! (n / 2)!
 Chap 06 Binomial Theorem 471

Sol. Given, (1 + x )n = C 0 + C 1x + C 2 x 2 +... + C n - 2 x n - 2 On multiplying Eqs. (i) and (ii), then we get
n (1 + x )2n - 1 = ( C 1 + 2C 2 x + 3C 3 x 2 +... + n C n x n -1 )
+ Cn - 1 x n - 1 + Cn x n …(i)
and (1 - x )n = C 0 - C 1x + C 2 x 2 - ... + ( - 1)n C n x n …(ii) ´ (C 0 x n + C 1 x n - 1 + C 2 x n - 2 + C 3 x n - 3 +... + C n ) …(iii)

On multiplying Eqs. (i) and (ii), we get Now, coefficient of x n - 1 on RHS

(1 - x 2 )n = ( C 0 + C 1x + C 2 x 2 +... + C n - 2 x n - 2 = C 12 + 2C 22 + 3C 32 + ... + n C n2
+ C n - 1 x n - 1 + C n x n ) ´ (C 2 - C 1x + C 2 x 2
- and coefficient of x n - 1 on LHS
... + ( - 1)n C n x n ) ...(iii) (2n - 1) !
= n × 2n - 1C n - 1 = n ×
( n - 1) ! n !
Now, coefficient of x n in RHS
( 2n - 1) ! ( 2n - 1) !
= C 0C n - C 1 C n - 1 + C 2 C n - 2 - K + ( - 1)n C n C 0 = =
(n - 1) ! (n - 1) ! {(n - 1) !)2 }
Now, general term in LHS,
But Eq. (iii) is an identity, therefore the coefficient of x n - 1
Tr + 1 = n C r ( - x 2 )r = ( - 1)r × n C r x 2 r
in RHS = coefficient of x n -1 in LHS.
Putting 2r = n , we get
2 2 2 2 (2n - 1) !
r = n /2 Þ C 1 + 2C 2 + 3C 3 + ... + nC n =
{(n - 1) !}
Now, Tn / 2 + 1 = ( -1)n / 2 × n C n / 2 x n
\ Coefficient of x n in LHS = ( -1)n / 2 . n C n / 2 y Example 79. If (1 + x )n = C 0 + C 1 x + C 2 x 2 +...+C n x n ,
n! C2 C2 C2 (2n + 1)!
= ( - 1)n / 2 × prove that C 0 2 + 1 + 2 + ... + n = .
( n / 2) ! ( n / 2) ! 2 3 n + 1 {(n + 1)!} 2
ì0 if n is odd
Sol. Given, (1 + x )n = C 0 + C 1 x + C 2 x 2 +... + C n x n
ï
=í n !
( -1)n / 2 × , if n is even Integrating both sides w.r.t. x within limits 0 to x, then we
ï ( n / 2) ! ( n / 2) !
î get
But Eq. (iii) is an identity, therefore the coefficient of x n in x x
ò0 (1 + x ) ò0 ( C 0 + C 1 x + C 2 x
n 2
dx = + ... + C n x n ) dx
RHS = coefficient of x n in LHS.
Þ C 0C n - C 1C n - 1 + C 2 C n - 2 - ... + ( - 1)n C n C 0 (1 + x )n + 1 - 1 C x2 C x3 C xn + 1
= C 0x + 1 + 2 + ... + n …(i)
(1 + n ) 2 3 n +1
ì0, if n is odd
ï and ( x + 1)n = C 0 x n + C 1 x n - 1 + C 2 x n - 2 + ... + C n …(ii)
=í n!
( -1)n / 2 × , if n is even
ï ( n / 2 ) ! ( n / 2) !
î
Multiplying Eqs. (i) and (ii), we get
3. If each term is the product of two binomial
1
coefficient divided or multiplied by an integer, { (1 + x )2n + 1 - (1 + x )n }
then integrating or differentiating by preceeding ( n + 1)
method. Then, multiplying two series and æ C x2 C x3 C xn + 1 ö
= çC 0 x + 1 + 2 + ... + n ÷
equate the coefficient of suitable power of x on è 2 3 n +1 ø
both sides.

y Example 78. If (1 + x )n = C 0 + C 1 x + C 2 x 2 ´ (C 0 x n + C 1 x n - 1 + C 2 x n - 2 + ... + C n ) …(iii)

+ C 3 x 3 +... + C n x n , prove that Now, coefficient of x n - 1 in RHS of Eq. (iii)

2 2 2
(2n - 1)! = C0 +
2 C1 C C
+ 2 + ... + n
C 12 + 2 C 22 + 3 C 32 + ... + nC n2 = . 2 3 n +1
((n - 1)!) 2
2 3 and coefficient of x n + 1 in LHS of Eq. (iii)
Sol. Given, (1 + x )n = C 0 + C 1x + C 2 x + C3 x +... + C n x n
1
Differentiating both sides w.r.t. x, we get = { 2n + 1C n + 1 - 0}
( n + 1)
n (1 + x )n - 1 = 0 + C 1 + 2C 2 x + 3C 3 x 2 +... + n C n x n -1 1 ( 2n + 1) !
= ×
Þ n (1 + x )n - 1 = C 1 + 2C 2 x + 3C 3 x 2
+... + n C n x n -1 ...(i) ( n + 1) ( n + 1) ! n !
and ( x + 1)n = C 0 x n + C 1x n - 1 + C 2 x n - 2 + C 3 x n - 3 ( 2n + 1) ! ( 2n + 1) !
= =
+... + C n …(ii) (n + 1) ! (n + 1) ! {(n + 1) !} 2
472 Textbook of Algebra

But Eq. (iii) is an identity, therefore coefficient of x n + 1 in [C 0 (1 + x )2n - C 1 (1 + x )2n -1 + C 2 (1 + x )2n - 2

RHS of Eq. (iii) = coefficient of x n + 1 in LHS of Eq. (iii). - C 3 (1 + x )2n - 3 + ... + ( - 1)n C n × (1 + x )n ]
C 12 C 22 C2 (2n + 1) ! = Coefficient of x n in
Þ C 02 + + + ... + n =
2 3 n + 1 {(n + 1) !} 2 (1 + x )n [C 0 (1 + x )n - C 1 (1 + x )n - 1 + C 2 (1 + x )n - 2
- C 3 (1 + x )n - 3 + ... + ( -1)n C n × 1]

Binomial Inside Binomial = Coefficient of x n in (1 + x )n [ ((1 + x ) - 1)n ]

= Coefficient of x n in (1 + x )n × x n d
The upper suffices of binomial coefficients are different = Constant term in (1 + x )n = 1 = RHS
but lower suffices are same.
n
y Example 80. Evaluate S n +r Cn .
r =0 Sum of the Series
n
Case I When i and j are independent.
Sol. S n + rCn = n
Cn + n +1
Cn + n+2
C n + ... + 2n
Cn
r =0 In this summation, three types of terms occur, when
= Coefficient of x n in i < j , i = j and i > j,
n ì ö üï
ï æ
n n n
[ (1 + x )n + (1 + x )n + 1 + (1 + x )n + 2 + ... + ... + (1 + x )2n ] i.e., S S a i a j = S ía i ç S a j ÷ ý
i=0 j =0 i=0 ç ÷
é (1 + x )n [(1 + x )n +1 - 1] ù ïî è j = 0 ø ïþ
= Coefficient of x n in ê ú 2 2
ë (1 + x ) - 1 û n n æ n ö æ n ö
= S a i S a j = ç S a i ÷ or ç S a j ÷
= Coefficient of x n + 1 in [ (1 + x )2n + 1 - (1 + x )n ] i=0 j =0 çi = 0 ÷ çj =0 ÷
è ø è ø
2n + 1 2n + 1 2
= Cn + 1 - 0 = Cn n n æ n ö
Corollary I S S n C i n C j = ç S n C i ÷
i=0 j =0 ç ÷
y Example 81. If (1 + x )n = C 0 + C 1 x èi = 0 ø
+ C 2 x 2+ ... + C n x n , prove that = (2 n ) 2 = 2 2n
C 0 × 2n C n - C 1 × 2n -2 C n + C 2 × 2n - 4 C n - ... = 2n
y Example 83. If (1 + x )n = C 0 + C 1 x
2n 2n - 2 2n - 4
Sol. LHS = C 0 × Cn - C1 × Cn + C 2 × C n - ...
+ C 2 x 2 +... + C n x n , find the values of the following.
n n
= Coefficient of x n in (i) S S (C i + C j )
i =0 j =0
[ C 0 (1 + x )2n - C 1 (1 + x )2n - 2 + C 2 (1 + x )2n - 4 - ... ]
n n
= Coefficient of x n in (ii) S S (i + j ) C i C j
i =0 j =0
[C 0 (1 + x )2 ] n - C 1 [(1 + x )2 ] n - 1 + C 2 [ (1 + x )2 ] n - 2 - ...]
n n n n n n
= Coefficient of x n in [[ (1 + x )2 - 1]n ] Sol. (i) S S (C i + C j ) = i S= 0 j S= 0 C i + i S= 0 j S= 0
i =0 j =0
Cj
n 2 n
= Coefficient of x in (2x + x ) n æ n ö n æ n ö
= Constant term in (2 + x ) = 2 = RHS n n = S ç S C i ÷÷ + i S= 0 çç j S= 0 C j ÷÷
j = 0 çi = 0
è ø è ø
n 2
y Example 82. If (1 + x ) = C 0 + C 1 x + C 2 x n n

3 n
= S (2n ) + S (2n ) = (n + 1) × 2n + (n + 1) × 2n
+ C 3 x +... + C n x , prove that j =0 i =0

= 2 (n + 1) 2n = (n + 1) 2n +1
C 0 × 2n C n - C 1 × 2n -1C n + C 2 × 2n - 2
Cn - C 3 × 2n - 3
C n + ... +
n n n n n n
( - 1)n C n × n C n = 1 (ii) S S (i + j )C i C j
i =0 j =0
= S S i Ci C j
i =0 j =0
+ S S j Ci C j
i =0 j =0
2n -1 2n - 2
Sol. LHS = C 0 × 2n
Cn - C1 × Cn + C 2 × C n - C 3 × 2n - 3 C n n æ n ö n æ n ö
+ ... + ( - 1)n C n × nC n = S i C i çç j S= 0 C j ÷÷ + j S= 0 j C j ççi S= 0 C i ÷÷
i =0
= Coefficient of x n in è ø è ø
 Chap 06 Binomial Theorem 473

=
n
S i C i (2 n ) + j S= 0 j C j (2 n )
n
(vii) SS (i × j ) C i C j
i =0 0 £i < j £n
n n æ n n ö
= 2n S i n C i + 2n j S= 0 j n C j çS S C i ÷÷ - S S C i
i =0 çi = 0 j =0
è ø i = j
n n Sol. (i) SS Ci =
2
= 2n S i × × n - 1C i - 1 + 2 n S
n n 0 £i < j £n
j × × n - 1C j - 1
i =0 i j =0 j n n

n n ( n + 1) S C i - i S= 0 C i
= n × 2n S n - 1C i - 1 + n × 2 n j S= 0 n - 1C j - 1
i =0
=
i =0
2
= n × 2n - 1
n -1
= n × 2 n × 2n -1 + n × 2 n × 2 n - 1
(ii) SS j Ci = S
r =0
n
Cr
2n - 1 2n 0 £i < j £n
= n × 2× 2 = n ×2
{(r + 1) + (r + 2) + (r + 3) + ... + n }
Case II When i and j are dependent. n -1
( n - r ) ( n + r + 1)
In this summation, when i < j is equal to the sum of the = S n
Cr ×
terms when i > j, if a i and a j are symmetrical. So, in this r =0 2
case n -1
n n =
1
S n C r (n 2 - r 2 + n - r )
S S ai a j
i=0 j =0
= SS ai a j + S S ai a j 2r=0
0 £i < j £n i=j n -1 n -1 n -1
(n + n ) S n C r - S r × n C r - S r 2 × nC r
1 2 1 1
+ SS ai a j =
2 r =0 2r=0 2 r=0
0 £ j <i £n
1 1
=2 SS ai a j + S S ai a j = (n 2 + n ) (2n - 1) - × n × (2n - 1 - 1)
2 2
0 £i < j £n i=j
1
n n - × n [(n - 1) (2n - 2 - 1) + 2n - 1 - 1]
S S ai a j - S S ai a j
i=0 j =0
2
Þ SS ai a j =
i=j = n ( 3n + 1) × 2 n-3

0 £i < j £n 2
When a i and a j are not symmetrical, we find the sum by Remark
Here, j and Ci are not symmetrical.
listing all the terms.
Corollary I
n n (iii) Here, i ¹ j i.e., i > j or i < j
S S n C i n C j - SS n
Ci × nC j But C i and C j are symmetrical.
SS
0 £i < j £n
n
Ci nC j =
i=0 j =0 i=j
\ S S Ci C j =2 SS Ci C j
2 i ¹ j 0 £i < j £n
n
(2 n ) 2 - S (n Ci )2
2 2n - 2n C n 2n !
æ 22n - 2n C n ö
=2ç
i=0 ÷ [from corollary I]
= = = 2 2n - 1 - è 2 ø
2 2 2 (n !) 2
= 2 2n - 2n
Cn
y Example 84. If (1 + x )n = C 0 + C 1 x (iv) SS Ci C j = SS Ci C j + S S Ci C j
+ C 2 x +... + C n x n , find the values of the following.
2 0 £i £ j £n 0 £i < j £n i = j

1 2n
(i) SS Ci (ii) SS j Ci =
2
(2 - 2n
Cn ) + 2n
C n [from corollary I]
0 £i < j £n 0 £i < j £n
1
= (22n + 2n
(iii) S S Ci C j (iv) SS Ci C j 2
Cn )
i¹j 0 £i £ j £n
(v) SS (C i ± C j )2 = SS (C i2 + C 2j ± 2 C i C j )
(v) SS (C i ± C j ) 2 0 £i < j £n 0 £i < j £n

0 £i < j £n = SS (C i2 + C 2j ) ± 2 SS Ci C j
0 £i < j £n 0 £i < j £n
(vi) SS (i + j ) C i C j
Q SS ( C i2 + C 2j )
0 £i < j £n
0 £i < j £n
474 Textbook of Algebra

n n
S S (C i2 + C 2j ) - 2 i S= 0 C i2
n
Let P= SS (i + j ) C i C j ...(i)
0 £i < j £n
i =0 j =0
=
2 Replacing i by n - i and j by n - j in Eq. (i), then we
n æ n n ö get
S ç S C i 2 + j S= 0 C 2j ÷÷ - 2 × 2nC n
i =0 ç j =0 P= SS (n - i + n - j ) C n - i C n - j
è ø 0 £i < j £n
=
2 [Q sum of binomial expansion does not
n
change if we replace r byn - r ]
S ( ( n + 1) C i +
2 2n
Cn ) - 2 × Cn 2n

=
i =0
P= SS (2n - i - j ) C i C j
2 0 £i < j £n
n n
[Q n C r = n C n - r ] …(ii)
( n + 1) S C i 2 + 2nC n i S= 0 1 - 2 × 2nC n
i =0
= On adding Eqs. (i) and (ii), we get
2
(n + 1) × 2n C n + 2n C n × (n + 1) - 2 × 2n
Cn 2P = 2n SS Ci C j
= 0 £i < j £n
2
= n × 2n C n or P =n SS Ci C j =
n 2n
(2 - 2n
Cn )
\ SS (C i ± C j ) = n × 2 2n
C n ± (2 2n
- 2n
Cn )
0 £i < j £n 2
0 £i < j £n [from corollary I]
[from corollary 1] (vii) SS (i × j ) C i C j = SS (i n C i ) ( j × n C j )
= ( n m 1) 2n
C n ± 22n ; SS (i + j )C i C j
0 £i < j £n 0 £i < j £n

0 £i < j £n = n2 SS n -1
Ci - 1 n -1
Cj -1
0 £i < j £n
Remark
SS ( Ci + Cj ) = n × 2n
= n2 ê
é 22 (n - 1) - 2n - 2 C n -1 ù
0£ i < j £ n ú [from corollary I]
êë 2 úû
(vi) SS (i + j )C i C j
æ 1 ö
0 £i < j £n = n 2 ç2 2n - 3 - × 2n - 2 C n - 1 ÷
è 2 ø
 Chap 06 Binomial Theorem 475

#L Exercise for Session 4

1. The coefficient of a4 b 8 c 9d 9 in the expansion of (abc + abd + acd + bcd )10 is
10 !
(a) 10 ! (b) (c) 2520 (d) None of these
4 !8 !9 !9 !
2. If (1 + 2x + 3x 2 )10 = a 0 + a1x + a 2 x 2 + ... + a 20 x 20, then a1 equals
(a) 210 (b) 20 (c) 10 (d) None of these

3. If (1 + x + x 2 + x 3 )5 = a 0 + a1x + a 2x 2
+ ... + a15 x 15, then a10 equals
(a) 99 (b) 100 (c) 101 (d) 110
4. Coefficient of x 15 3
in (1 + x + x + x ) is 4 n

5 5 5 3
(a) S nC5 - r × nC3r
r =0
(b) S n C5r
r =0
(c) S n C2r
r =0
(d) S n C3 - r × n C5r
r =0
n
æ 1ö
5. The number of terms in the expansion of ç x 2 + 1 + 2 ÷ , n Î N is
è x ø
n +2 n +3 2 n +1 3n + 1
(a) C2 (b) C2 (c) C2 n (d) C3 n

6. If (1+ x ) 10
= a 0 + a1x + a 2 x + ...+ a10 x , then (a 0 - a 2 + a4 - a 6 + a 8 - a10 ) + (a1 - a 3 + a 5 - a 7 + a 9 )2 is equal to
2 10 2

(a) 2 9 (b) 3 9 (c) 2 10 (d) 3 10

7. If (1+ x )n = C0 + C1x + C2 x 2 + C3 x 3
+ ...+ Cn x n, n being even the value of
C0 + (C0 + C1 ) + (C0 + C1 + C2 ) + ... + (C0 + C1 + C2 + ... + Cn - 1 ) is equal to
(a) n × 2n (b) n × 2n - 1 (c) n × 2n - 2 (d) n × 2n - 3
C0 C C C Cn
8. The value of - 1 + 2 - 3 + ... + ( - 1)n is
1× 3 2 × 3 3 × 3 4 × 3 (n + 1) × 3
3 n+1 1
(a) (b) (c) (d) None of these
n+ 1 3 3 (n + 1)
æ50ö æ50ö æ50ö æ50ö æ50ö æ50ö æn ö
9. The value of ç ÷ ç ÷ + ç ÷ ç ÷ + ... + ç ÷ ç ÷ , where nCr = ç ÷ , is
è 0ø è 1ø è 1ø è 2ø è49ø è50ø è rø
2
æ 100ö æ 100ö æ 50ö æ 50ö
(a) ç ÷ (b) ç ÷ (c) ç ÷ (d) ç ÷
è 50 ø è 51 ø è 25ø è 25ø

10. If Cr stands for 4Cr , then C0 C4 - C1 C3 + C2 C2 - C3 C1 + C4 C0 is equal to

(a) C1 (b) C2 (c) C3 (d) C4
n
11. The sum
r =0
S (r + 1) ( nCr )2 is equal to

(n + 2) (2n - 1) ! (n + 2) (2n + 1) ! (n + 2) (2n + 1) ! (n + 2) (2n - 1) !

(a) (b) (c) (d)
n ! (n - 1) ! n ! (n - 1) ! n ! (n + 1) ! n ! (n + 1) !
n æ r -1 n r ö
12. S ç S Cr Cp 2p ÷ is equal to
r =1 ç
èp = 0
÷
ø
(a) 4n - 3n + 1 (b) 4n - 3n - 1 (c) 4n - 3n + 2 (d) 4n - 3n

æ 10 10 ö æ 10 10
C ö
13. çç S Cr ÷÷ çç S ( - 1)m mm ÷÷ is equal to
èr = 0 ø è m = 0 2 ø
(a) 1 (b) 2 5 (c) 2 10 (d) 2 20

14. The value of SSSS 2 is equal to

0£i < j <k<l £n
n+1 n+ 2
(a) 2 (n + 1) 3 (b) 2 × C4 (c) 2 (n + 1)4 (d) 2 × C3
 Shortcuts and Important Results to Remember
n
1 (r + 1th
) term from end in the expansion of r +1 æ xö
( x + y )n = (r + 1th 13 If the coefficients of x r , x in the expansion of ç a + ÷
) term from beginning in the expansion of è bø
( y + x )n . are equal, then n = (r + 1) (ab + 1) - 1, where n, r Î N and
2 If n Cr - 1, nCr , nCr + 1 are in AP, then (n - 2 r )2 = n + 2 or a, b are constants.
n
æ b ö
1
r = (n ± (n + 2 )) for r = 2, n = 7 and for r = 5, n = 7, 14. 14 Coefficient of x m in the expansion of ç ax p + q ÷
2 è x ø
np - m
= Coefficient of Tr + 1, where r = , where p, q, n Î N
3 Four consecutive binomial coefficients can never be p+q
in AP. and a, b are constants.
4 Three consecutive binomial coefficients can never be in 15 The term independent of x in the expansion of
n
GP or HP. æ p b ö np
ç ax + q ÷ is Tr + 1, where r = , where n , p, q Î N
5 If a, b, c , d are four consecutive coefficients in the è x ø p +q
a b c
expansion of (1 + x )n , then , , are in AP. and a, b are constants.
a+ b b+c c +d
16 Sum of the coefficients in the expansion of (ax + by )n is
a c æ b ö
(i) + =2 ç ÷ (a + b)n , where n Î N and a, b are constants.
a+ b c +d èb + cø
2 17 If (1 + x )n = C0 + C1 x + C2 x 2 + ... + Cn x n and p + q = 1, then
æ b ö ac
(ii) ç ÷ > n
èb + cø (a + b) (c + d ) (i) S r × Cr × pr × q n - r
r =0
= np
2n
6 If greatest term in (1 + x ) has the greatest coefficient, n

then
n
<x<
n+1
.
(ii) S r 2 × Cr × pr × q n - r
r =0
= n 2 p2 + npq
n+1 n
18 If (1 + x )n = C0 + C1 x + C2 x 2
+ ... + Cn x n, then
7 (a) The coefficient of x n - 1 in the expansion of
2n + 1 - 1
n n
( x - 1) ( x - 2 ) ( x - 3) ... ( x - n ) = - (1 + 2 + 3 + ... + n ) (i) S r × Cr = n × 2 n -1 (ii) S Cr
=
n (n + 1) r =0 r =0 r +1 n+1
=- = - n + 1C2 n n
2
(iii) S r 2 × Cr = n (n + 1) 2 n - 2 (iv) S (-1)r × r × Cr =0
(b) The coefficient of x n - 1 in the expansion of r =0 r =0
n r
( x + 1) ( x + 2 ) ( x + 3) K ( x + n )
n (n + 1) n + 1 (v) S (-1) Cr =
1
= (1 + 2 + 3 + K + n ) = = C2 r =0 r +1 n+1
2 n
8 The number of terms in the expansion of (vi) S (-1)r Cr
r =0
= 1+
1 1
+ +K+
1
r 2 3 n
ìn + 2
ïï 2 , if n is even n
n
( x + a) + ( x - a) = ín
(vii) S (- 1)r × r 2 × Cr =0
ï n + 1 , if n is odd r =0
ïî 2 n

9 The number of terms in the expansion of (viii) S (- 1)r × (a - r ) (b - r ) Cr

r =0
= 0, " n > 3
ìn n
ïï 2 , if n is even
n
( x + a) - ( x - a) = ín (ix) S (- 1)r (a - r ) (b - r ) (c - r ) Cr
r =0
= 0, " n > 3
ï n + 1 , if n is odd n
ïî 2
(x) S (- 1)r (a - r )3 Cr
r =0
= 0, " n > 3
10 The number of terms in the expansion of multinomial
n k
( x1 + x2 + x3 + ... + xm )n , when x1, x2 , x3 , ... , xm Î C and
(xi) S r(r - 1) (r - 2 )... (r - k + 1) Cr x r -k = d k (1 + x )n
n Î N, is n + m - 1Cm - 1. r =0 dx
n
d2
11 The number of terms in the expansion of for k = 2 , S r (r - 1) Cr = 2 [(1 + x )n ]x = 1 = n(n - 1) 2 n - 2
n r =0 dx
æ p b ö
ç ax + + c ÷ , where n, p Î N and a, b, c are n
è xp ø and for k = 3; S r(r - 1) (r - 2 ) (-1)r - 3Cr
r =0
constants, is 2n + 1.
d3
12 If the coefficients of pth and qth terms in the expansion of = [(1 + x )n ]x = - 1 = 0
(1 + x )n are equal, then p + q = n + 2, where p, q, n Î N. dx 3