Binomial Theorem

INTRODUCTION
Binomial is made of “Bi” which means “two”
and “nomial” which means an expression with
numbers or variables.

Binomial Expression Definition

1 An algebraic expression which
eg : x + y, yz +
x contains two dissimilar terms is
called binomial expression.
Note :
 x + (y + z) is also a binomial expression
where ‘x’ is first term and ‘(y + z)’ is second
term. Similarly, (x + y) + (z + w) is a binomial
expression where (y +z) is first term and
(z + w) is second term.

Binomial Theorem
(x + y)2 = x2 + y2 + 2xy Definition
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 The formula by which any
positive integral power of a
When index is very high then use binomial binomial expression can be
theorem as below expanded in the form of a series
(x + y)n = nC0xny0 + nC1xn–1y1 + nC2xn–2y2 + … + is known as Binomial Theorem.
nCrxn–ryr+ … + nCnxn–nyn
Where n∈N, x and y ∈ set of complex
numbers and nC0, nC1, nC2… are called binomial
coefficients normally written as C0, C1, C2…
(n hidden)
Point to Remember!!!
Proof of Binomial Expansion
(x + y)n = (x + y) (x + y)…(x + y) {n brackets} (i) The number of terms in the
Each term in the expansion is formed by taking expansion is (n + 1)
one letter from each bracket and multiplying i.e. one more than the index
them together. For example, choose x from (n)
all the bracket and multiplying them, xn is (ii) Tr+1 = nCrxn–ryr is called the
obtained. This can be done in 1 (= nC0) way. general term of the binomial
expansion.
Binomial Theorem

Now, choose x from (n – 1) brackets and y

from remaining one bracket and multiplying (iii) Sum of powers of each term
in (x + y)n is n.
them, xn–1y1 is obtained. But number of ways
of choosing (n – 1) brackets out of ‘n’ is
n
Cn–1 = nC1 ways.
1.
 Hence nC1 × 1C1 = nC1 = nC1 terms of the form
xn–1y1 are generated. Continuing in this way, nCr
terms of the form xn–ryr are generated & the
Point to Remember!!!
process is done till all possible combinations
of powers of ‘x’ and ‘y’ are obtained. Adding
all (x + y)n = xn + nC1xn–1y1 +…+ nCrxn–ryr+…+ yn Number of distinct terms in the
expansion of (x1+x2+…+xr)n is
Highlights
n+r–1
Cr–1 . Which is also equal to
n n number of ways of distributing
( )
n
x+y = ∑  n Cr xn−r yr = ∑T r+1
‘n’ identical coins among ‘r’
r= 0 r= 0 persons.
Where Tr+1 is general term
n

( ) ( )
n n
∑  C y
n n−r r
Also, x + y = y+x = r
x
r= 0

Q. Find number of distinct terms in the expansion of

(a) (x + y + z)2 (b) (x + y + z)8

Sol.
(a) n = 2, r = 3
n+r–1
Cr–1 = 3+2–1C3–1 = 4C2 = 6
(b) n = 8 r = 3
n+r–1
Cr–1 = 8+3–1C3–1 = 10C2 = 45

Sum of all binomial coefficients in expansion of (x + y)n

C0 + C1 + C2 +…+ Cn = 2n …(1)

Proof :
Put x = y = 1 in expansion of (x + y)n
(x + y)n = nC0xny0 + nC1xn–1y1 + nC2xn–2y2+…+ nCnxn–nyn
(1 + 1)n = nC0 1n10 + nC1 1n-1 11 +…+ nCn 10 1n
2n = nC0 + nC1 +…+ nCn.
Note :
(x + 2y)2 = 2C0x2–0(2y)0 + 2C1x2–1(2y)1 + 2C2x2–2(2y)2
= 1x2 + 4xy + 4y2
y Here, 2C0, 2C1, 2C2 are binomial coefficients and 1, 4, 4
Binomial Theorem

are coefficients.
y To find sum of binomial coefficients in expansion
of (x + 2y)2, use formula (1) and to find sum of all
coefficients in the expansion of (x+2y)2 put x = y = 1,
i.e. [1 + 2(1)]2 = 9
2.
 Q. Find the sum of all coefficients in the expansion of
(a) (x – y + z)2 (b) (2x2 – 3x + 3)2021
(c) (x + 3)2020 (d) (x – 1 )99

Sol.
(a) put x = y = z = 1
(1 – 1 + 1)2 = 1
(b) put x = 1, [2(1)2 – 3(1) + 3]2021 = 22021
(c) put x = 1, (1 + 3)2020 = 42020
(d) put x = 1, (1 – 1)99 = 0

pth term from the end

y pth term from the end in the expansion of
(x + y)n is same as (n – p + 2)th term from the
beginning of the expansion
y rth term binomial coefficient from the end is
n
Cn–(r–1) = nCn–r+1
y (x + y)n = nC0xny0 + nC1xn–1y1 + nC2xn–2y2 + …
+ nCn–2x2yn–2 + nCn-1x1yn–1 + nCnx0yn
Binomial coefficients of terms equidistance
from beginning and end are equal i.e.
n
C0 = nCn
n
C1 = nCn–1
n
Cr = nCn–r

Frequently Used Binomial Expansion

(a) (x + y)n = nC0xny0 + nC1xn–1y1+…+ nCnx0yn Point to Remember!!!
n

∑  C x
n n−r r
= r
y
r= 0 (1) Coefficient of xr in expansion
of (1 + x)n = nCr
(b) (x – y)n = nC0xny0 – nC1xn–1y1 + nC2xn–2y2 –
nC3xn–3y3+…+ (–1)n nCnx0yn (2) Coefficient of xr in expansion
n of (1 – x)n = (–1)r nCr
∑ ( −1)
r n
= Cr xn−r yr
r= 0

(c) (1 + x)n = nC0 + nC1x + nC2x2+…+ nCrxr +…+ nCnxn

n

∑  C x
Binomial Theorem

n r
= r
r= 0

(d) (1 – x) = nC0 – nC1x + nC2x2 – nC3x3 + …

n

+ (–1)r nCrxr +…+ (–1)n nCnxn

n

∑ ( −1)
r n
= Cr xr
r =0
3.
 Q. Find coefficient of x6 in (1 + 3x + 3x2 + x3)15

Sol. Given expansion is [(1 + x)3]15 = (1 + x)45

∴ coefficient of x6 in (1 + x)45 is 45C6

Q. Show that coefficient of x5 in the expansion of (1 + x2)5. (1 + x)4 is 60.

Sol. (5C0 + 5C1x2 + 5C2x4 + 5C3x6 + 5C4x8 + 5C5x10) × (4C0 + 4C1x + 4C2x2 + 4C3x3 + 4C4x4)
Required coefficient is
5
C1 4C3 + 5C2 4C1 = 20 + 40 = 60.

Q. n+ 4

( ) . (1 + x) = ∑a x . If a , a
2 n
Let 1 + x2 k
k
1 2
and a3 are in A.P. find n.
k =0

Sol.
(1 + 2x2 + x4) (1 + nC1x + nC2x2+…) =
a0x0 + a1x1 + a2x2 +…
comparing coefficients of like powers of x both side
a1 = nC1, a2 = 2 + nC2, a3 = nC3 + 2(nC1)
2a2 = a1 + a3


2 2 +
n n−1 ( )  = nC +
(
n n−1 n−2 )( ) + 2n
1
 2  6
 
6[4 + n(n – 1)] = 18n + n3 – 3n2 + 2n
24 + 6n2 – 6n = 18n + n3 – 3n2 + 2n

n3 – 9n2 + 26n – 24 = 0
(n – 2) (n – 3) (n – 4) = 0 ⇒ n = 2, 3, 4

Q. The sum of coefficients of integral powers of x in the binomial expansion of

(1 − 2 x )
50
is

(A)
1 50
2
(
2 + 1 (B)
1 50
2
)
3 +1 ( ) (C)
2
( )
1 50
3 (D)
2
(
1 50
)
3 −1

Ans. (B)
Binomial Theorem

(1 − 2 x ) = ( ) + C (2 x ) − C (2 x )
50 1 2 3

Sol.
50
C0 − 50C1 2 x 50
2
50
3

+ C (2 x ) (2 x ) + … + (−1) C (2 x ) …(i)
4 5 50
50 50 50
4
− 50C5 50

4.
 Required sum = C0 +
50
C2 22 +
50
C424 +…+
50
C50250
50

(1 + 2 x )50 = 50C0 + 50C1 (2 x )1 + 50C2 (2 x )2 + 50C3 (2 x )3 + … + 50C50 (2 x )50

…(ii)
(i) + (ii) gives

( ) ( 
) ( ) ( ) ( )
50 50 2 4 50 
1−2 x + 1+ 2 x = 2  50C0 + 50C2 2 x + 50C4 2 x + ... + 50C50 2 x 
 
put x = 1 both side
( 1 − 2) ( )
50 50
+ 1+ 2
= 50
C0 + 50
C222 + C424 +…+
50
C50250
50
2

On solving L.H.S. is
1 50
2
(
3 +1 . )
Q. Sum of all coefficients in (1 + x – 3x2)2163

Sol. Put x = 1
(1 + 1 – 3)2163 = – 1

Q. The sum of the coefficients in the expansion of (a + 2b + c)10 is

(A) 410 (B) 310 (C) 210 (D) 104

Ans. (A)

Sol. Put a = b = c = 1
(1 + 2(1) + 1)10 = 410

Q.  y
7

Find the fourth term in the expansion of   2x − 

 2

3
 y
Sol. T4 = T3+ 1 = 7C3 (2x)7 −3  − 
 2
7×6×5 y3
= × 16x4 ( −1)
6 8
 = – 70x4y3
Binomial Theorem

5.
 6
  1  
Q.   1+log 10x  1

If the fourth term in the binomial expansion of  x + x  is equal to
12

 
 
200, and x > 1, then the value of x is
(A) 103 (B) 100 (C) 104 (D) 10

Ans. (D)
3
 1
  1   2   1 
3
  1+log 10x   
Sol.
6
C3   x  
 x 12  = 200
  
  
  
 
3 1 
1
 
2  1+log 10 x 
x . x 4 = 10

3 1  1
  log 10 x + log 10 x = 1 (Taking log both side)
2  1 + log 10 x  4
3 1 α
α + = 1 ⇒ α2 + 3α – 4 = 0
(
2 1+ α 4 )
α = 1, – 4
∴ log10x = α = 1, – 4
x = 10, 10–4

Q.  1 
6

Find term involving x in  2x2 − 

3

 3x 

( )  1 
6 −r

Sol.
6 2
Tr + 1 = Cr 2x − 
 3x 
r
x 12−2r  1 
= 6Cr 26−r ( ) − 
xr  3 

Equating power of x to 3
Binomial Theorem

12 – 3r = 3 ⇒ r = 3
3

( )  1  −160 3
3

∴ Term = 6C3 2x2 −

 3x
 =
 27
x

6.
 Q.
The coefficient of (3r)th term and coefficient of (r + 2)th term in the expansion
of (1 + x)2n are equal (r > 1, n > 2 and positive integer) then
n n n+ 1 n−1
(A) r = (B) r = (C) r = (D) r =
2 3 2 2

Ans. (A)

Sol. C3r–1 =
2n 2n
Cr+1

3r − 1 = r + 1 or 2n = 3r − 1 + r + 1   n C = nC 
x y
 
2r = 2 or 2n = 4r           ⇒ x + y = n or 
 
r=1 or n = 2r  x=y 
 

Q. If in the expansion of (1 + y)n, then coefficient of 5th, 6th and 7th terms are in
A.P., then ‘n’ is equal to
(A) 7, 11 (B) 7, 14 (C) 8, 16 (D) None

Ans. (B)

Sol. 2 nC5 = nC4 + nC6

2 n!( ) =
n!
+
n!
(
5! n − 5 ! ) (
4! n − 4 ! ) (
6! n − 6 ! )
12(n – 4) = 30 + (n – 5) (n – 4)
n2 – 21n + 98 = 0 ⇒ n = 7, 14

Q. 2n 2n

∑a ( x − 2) = ∑b ( x − 3)
r r
If r r
and ak = 1 ∀ k ≥ n, then show that bn = 2n+1Cn+1
r= 0 r= 0

Sol.
a0(x – 2)0 + a1(x – 2)1 + a2(x – 2)2 +…+ an–1(x – 2)n–1 + an(x – 2)n + an+1(x –2)n+ 1+…+
a2n(x – 2)2n =
b0(x – 3)0 + b1(x – 3)1 + b2(x – 3)2 +…+ b2n(x – 3)2n

a0 + a1(x – 2)1 + a2(x – 2)2 +…+ an–1(x – 2)n–1 + (x –2)n + (x – 2)n+1 +…+ (x – 2)2n
= b0 + b1(x – 3)1 + b2(x–3)2 +…+ bn(x – 3)n + bn+1(x – 3)n+1 +…+ b2n(x – 3)2n
(M 1)
Comparing coefficient of xn from both sides-
Binomial Theorem

n
C0 + n+1C1 + n+2C2 +…+ 2nCn = bn
bn = (n+1Cn+1 + n+1Cn) + n+2Cn +…+ 2nCn
= (n+2Cn+1 + n+2Cn)+…+ 2nCn

7.
 [nCr + nCr+1 = n+1Cr+1]
= ((n+3)Cn+1 + n+3Cn) +…+ 2n
Cn

= 2n+1Cn+1
(M 2)
(x + 1)n + (x + 1)n+1 + (x + 1)n+2 +…+ (x + 1)2n = S(say)
To find coefficient of xn from above series.
A = (x + 1)n R = (x + 1)
 
( )
n+ 1
 Rn − 1  n  x + 1 − 1
S=A   = x+1 ( )
 R − 1  
x+1 −1 
 ( )
( x + 1) ( ) ( x + 1) ( x + 1)
2n+ 1 n 2n+ 1 n
− x+1
S= = −
x x x
∴ coefficient of xn in S = 2n+1
Cn+1 (coefficient of xn+1 in (1 + x)2n+1)
∴ bn = 2n+1Cn+1

Term Independent of x
Step-I: Write general term of the expansion.
Step-II: Equate the exponent of x to zero

Q.  1
6

Find the term independent of x in  2x2 −

 3x 

( )  1 
6 −r

Sol. Tr + 1 = 6Cr 2x2 − 

 3x 
r
6 6 −r 12−2r  1 1
  = Cr 2 x −  ⋅ r
 3 x
r
 1
  = Cr 2  −  x 12−3r
6 6 −r

 3
For term independent of x, 12 – 3r = 0
r=4
Hence T4+1 = T5 or 5 term is independent of x
th

4
 1
Binomial Theorem

T5 = 6C4 26−4  −  x0
 3
1 20
= 15 × 4 × =
81 27

8.
Middle Term
The middle term in the expansion of (x + y)n
depends upon the value of n
(i) If n is even
Total number of terms in the expansion
= n + 1 (odd)
There is only one middle term
th
n 
 + 1 term is the middle term
2 
(ii) If n is odd
Total number of terms in the expansion
= n + 1 (even)
There is two middle term
th th
n + 1 n + 1 
  term and  2 + 1 term are two
 2 
middle term.

Q.  1 
11

Find middle terms in the expansion of  2x2 − 

 3x 

n+ 1 n+ 1
Sol. n = 11
2
= 6
2
+1=7

T6 and T7 are middle terms

5 6

( )  1 
( )  1 
6 5
T6 = − 11C5 2x2   ; T7 = 11C6 2x2 − 
 3x   3x 
9856 7 4928 4
=− x ; = x
81 243
Note :
(i) Middle term has greatest binomial coefficient and if there are two
middle terms their binomial coefficients will be equal

n
When r = if n ∈ even
Binomial Theorem

2
(ii) nCr will be maximum
n−1 n+ 1
When r = or if n ∈ odd
2 2

9.
 Q. Find the coefficient of x15 in the expansion of (x – x2)10.

Sol.
[x(1 – x)]10 = x10 (1 – x)10
∴ coeff. of x15 in x10(1 – x)10 =
coeff. of x5 in (1 – x)10 = (-1)5 10C5
= – 252

 3200 
Q.
If {P} denotes the fractional part of the number P, then 
 8 
 , is equal to

5 1 7 3
(A) (B) (C) (D)
8 8 8 8

Ans. (B)

Sol. 3200 = (1 + 8)100 = 1 + 100C1 81 + 100C2 82 +…

= 1 + 8k, k ∈ integer
 3200   1 + 8k   1   1 1
∴  =  =  + k =   =
 8   8   8  8  8

( ) + (1 − 2 )
7 7

Q. Value of 1 + 2

(A) 402 (B) 458 (C) 478 (D) 498

Ans. (C)

(1 + 2 ) ( 2 ) + 7C ( 2 ) ( 2)
7 1 2 3

Sol. = 1 + 7C1 2
+ 7C3 +…+

(1 − 2 ) = 1 − 7C ( 2 ) + 7C ( 2 ) ( 2)
7 1 2 3
1 2
− 7C3 + ... 

( ) ( )
7 7
1+ 2 + 1− 2
 ( )
= 2  1 + 7C2 2 + 7C4 22 + 7C6 23 

= 2[1 + 42 + 140 + 56]
= 478

( ) −( )
5 5

Q. Find value of 3+3 3 −3

Binomial Theorem

( ) ( )
5  5 5

Sol. 3  1+ 3

− 1− 3 


10.
 ( ) ( 3) + C ( 3) ( 3 ) 
5  5 1
5
3 5 

= 3 2  C 1 3
+ 5C5
 
= 54[5 + 30 + 9]
= 54 × 44 = 2376

Q. In the expansion of (1 + x)10, coefficients of (4r + 5)th term = coefficient of

(2r + 1)th term. Find r

Sol.
C4r + 4 = 10C2r
10

4r + 4 = 2r or 10 = (4r + 4) + 2r ⇒ r = 1
2r = – 4 (not possible)
[nCx = nCy ⇒ x + y = n or x = y]

Q. Let α > 0, β > 0 be such that α3 + β2 = 4. If the maximum value of the term
10
 1 −1 
independent of x in the binomial expansion of  αx 9 + βx 6  is 10k, then k is
 
 
equal to
(A) 176 (B) 336 (C) 352 (D) 84

Ans. (B)
10 −r r
 1  −1 
Sol.
10
Tr + 1 =   Cr  αx 9   βx 6 
   
   
10 − r r
For term independent of x, − =0 ⇒r=4
9 6
∴ T5 = 10C4 α6β4
1
α3 + β2

2
≥ α3β2 ( ) 2

2
4 3 2
  ≥ α β ⇒ (α β )max = 4
3 2
2
 
∴ (T5)max = 10 k
Binomial Theorem

10
C4(4)2 = 10 k
k = 336

11.
 Q.  cosα 
10
Greatest value of term independent of x in  xsinα + x ∈R
 x 

r
 cos α 
Sol.
10
Tr + 1 = Cr (x sin α )10−r 
 x 

For term independent of x, 10 – 2r = 0 ⇒ r = 5

T6 = 10C5(sinα)5 (cosα)5
10 10
C5 C5
= 5
× (2 sin α sin α )5 = 5
(sin 2α )5
2 2
10
C5
( T6 )max = 25

Q.  1
n

Sum of all coefficient in the expansion of  x +  is 4096. Find greatest

 x
binomial coefficients.

Sol. Put x = 1
(1 + 1)n = 4096 = 212 ⇒ n = 12
Greatest binomial coefficient = 12C6

( )
15

Q. In the expansion of 2+ 7 , number of rational terms are:

( ) ( )
15−r r

Sol.
15
Tr + 1 = Cr 2 7

For rational terms 15 – r and r should be even which is not possible

Hence, all terms will be irrational
So, number of rational terms is zero.
Binomial Theorem

12.
 Q.  1 
20

 43 + 1  . Find number of rational terms.

 1 
 
 64 

20 −r r
 1  −1 
Sol.
20
Tr + 1 = Cr  4 3  6 4 
   
   
For rational terms, ‘r’ should be multiple of 4
∴ Possible values of r is 0, 4, 8, 12, 16, 20
20 − r
but at r = 8 and 20 only is integer
3
∴ r = 8,20
Hence, number of rational terms are 2

Q. In the expansion of y = 1 + (1 + x) + (1 + x)2 +…+ (1 + x)19, if the coefficient of xp

is greatest. Find p

(1 + x) − 1 = (1 + x)
20 20
−1
Sol. y =
(1 + x) − 1 x

∴ coefficient of xp in y is 20Cp+1
20
Cp+1 is greatest when p + 1 = 10 ⇒ p = 9.

Q. Show that C02 + C21 + C22 + C23 + ... + Cn2 = 2nCn

(M 1)
By method of selection.
Consider n boys and n girls in a class. Total number of ways to select n
students out of 2n is 2nCn. Alternatively, the same selection can be done if out
of n boys, zero boys and out of n girls, n girls are selected or out of n boys, 1
boy and out of n girls, (n–1) girls are selected and so on…
Hence 2n
Cn = nC0 nCn + nC1 nCn–1 + nC2 nCn–2 +…
= nC0 nC0 + nC1 nC1 + nC2 nC2 +…
∴ 2nCn = C02 + C21 + C22 + ...
Binomial Theorem

(M 2) Multiplication of two series

(1 + x)n = C0 + C1x + C2x2 + …
1
Replace x by
x

13.
 n
 1 C 1 C2
 1 +  = C0 + + 2 + ...
 x x x
n
 C1 C2 



1
x
(
(1 + x)n  1 +  = C0 + C1x + C2 x2 + … )  C 0
+
x
+
x2
+ …



(1 + x)
2n
 C C   C C   C C 
= C0 C0 + 1 + 22 + ... + C1x C0 + 1 + 2 + ... + C2 x2 C0 + 1 + 22 + ... + …
xn  x x   x x2   x x 
Taking coefficient of x0 from both side
coefficient of xn in (1 + x)2n = C02 + C21 + C22 + ... + Cn2

Cn = C02 + C21 + C22 + ... + Cn2

2n

Q. Find S where S = 1.C02 + 2.C21 + 3.C22 + ... + n + 1 Cn2 ( )

Sol. S = 1.C02 + 2.C21 + 3.C22 + ... + n + 1 Cn2 ( )

( )
+S = n + 1 Cn2 + nCn2− 1 + ... + 1 . C02

( ) ( )
2S = n + 2 C02 + n + 2 C21 + ... + n + 2 Cn2 ( )
S=
(n + 2) × 2nC
n
2

Q. Prove that (72C36 – 1) is divisible by 73.

Sol.
C36 = 73C36 – 72C35 (nCr+1 = n+1Cr+1 – nCr)
72

  = 73C36 – (73C35 – 72C34)

= 73C36 – 73C35 + (73C34 – 72C33)
72
C36–1 = 73C36 – 73C35 + 73C34 – 73C33 + 73C32 – … – 73C1 + 72C0 – 1]

= [73C36 – 73C35 + 73C34 – 73C33 + 73C32– … – 73C1]

Each of the term contains a factor 73.
Hence, (72C36–1) is divisible by 73.
Binomial Theorem

14.
Product Of Two Binomial

Q. Show that C0C1 + C1C2 + C2C3 + … + Cn–1 Cn = 2nCn+1

Sol. (M-1)
(1 + x)n = C0 + C1x + C2x2 + C3x3 +…
n
 1  1  1   1 
 1 +  = C0 + C1   + C2  2  + C3  3  + ...
 x x x  x 
n
 C1 C2 



1
x
(
(1 + x)  1 +  = C0 + C1x + C2 x2 + … .
n
)  C
0
+
x
+
x2
+ …


1
Equating coefficients of both sides-
x

( )
n
1 1+ x
Coefficient of in (1 + x)n = C0C1 + C1C2 + C2C3+…+ Cn–1Cn
x xn
Coefficient of xn–1 in (1 + x)2n = C0C1 + C1C2 + C2C3 +…2nCn–1 = Cn+1
2n

So, C0C1 + C1C2 + C2C3 +… = 2nCn+1

(M-2)
LHS = C0Cn–1 + C1Cn–2 + C2Cn–3 +…+ Cn–1C0 consider ‘n’ boys and ‘n’ girls out of
which ‘n–1’ students are to be selected. First way is to select zero boys out of
n and (n – 1) girls out of n. For this number of ways are nC0 . nCn–1 second ways
is to select 1 boy out of n and (n – 2) girls out of n. This can be done in nC1 nCn–2
ways and so on…
Hence total number of ways equals L.H.S. Now, another ways is to directly
select (n – 1) students out of ‘2n’ in 2nCn–1 ways which is equal to 2nCn+1 (R.H.S)
Hence, C0C1 + C1C2 +… = 2nCn+1

Q. For r = 0, 1, …,10 Let Ar, Br and Cr denote respectively, the coefficient of xr in

10
the expansion of (1 + x)10, (1 + x)20 and (1 + x)30. Then ∑A (B
r=1
r 10 r )
B − C10 . Ar is
equal to
(A) B10 – C10 (B) A10(B102 – C10A10) (C) 0 (D) C10 – B10

Ans.
Binomial Theorem

(D)

Sol. Ar = 10Cr, Br = 20
Cr, Cr = 30
Cr
10 10

∑( ∑(
10
B10 Cr )(20 Cr ) − C10 10
Cr )2
r =1 r =1

15.
  10   10 
20
∑
C10   10 C10−r  20 Cr  −
 r = 0 
30
∑
C10  ( 10 Cr )2  −

r =0

20
C10 + 30
C10

20
C10 30C10 – 30C10 20C10 + 30
C10 – C10
20

30
C10 – 20C10 = C10 – B10

Q. The value of r for which 20Cr 20C0 + 20Cr–1 20C1 + 20Cr–2 20C2 +…+ 20C020Cr is
maximum is
(A) 15 (B) 20 (C) 11 (D) 10

Ans. (B)

Sol. Out of 20 boys and 20 girls, total ‘r’ students are selected. This is stated by the
given series.

Hence it equals Cr = 40Cr

40
(20+20)

40
Cr is maximum at r = = 20
2

Q. Show that C0 + 2C1 + 3C2 + 4C3 +…+ (n + 1) Cn = (n + 2) . 2n–1

Sol. (M 1)
S = C0 + 2C1 + 3C2 + 4C3 +…+ (n + 1)Cn
+ S = (n + 1) Cn + nCn–1 +…+ 2C1 + C0
2S = (n + 2) [C0 + C1 +…+ Cn]
n+2 n
S= [2 ]= (n + 2) 2n–1
2

(M 2)
Consider
(1 + x)n = C0 + C1x + C2x2 +…+ Cnxn
d   d
( )
n
1+ x x = [C0x + C1x2 + C2x3 + … + Cnxn+1]
dx   dx
(1 + x)n + nx(1 + x)n–1 = C0 + 2C1x + 3C2x2 + … + (n + 1)Cnxn
Put x = 1
2n + n2n–1 = C0 + 2C1 + 3C2 + … + (n + 1)Cn
Binomial Theorem

∴ C0 + 2C1 + 3C2 + … + (n + 1)Cn = (n + 2) 2n–1

16.
 Q. Show that S = 1.C1 + 2.C2 + 3.C3 +…+ n.Cn = n.2n–1

Sol. (M 1)
Consider
(1 + x)
= C0 + C1x + C2x2 +…+ Cnxn
n

Differentiate both sides with respect to x

n(1 + x)n–1 = C1 + 2C2x +…+ n.Cnxn–1
Put x = 1
n(2)n–1 = C1 + 2C2 + 3C3+…+ n.Cn
(M 2)
Tr = r.nCr


= n.n–1Cr–1   n Cr =

n n− 1
r

  Cr − 1 

( )
n n
S = n.
∑
r=1
n− 1
Cr − 1 = n ∑ 
r=1
n− 1
Cr − 1

= n.(2n–1)

Q. Which is larger: (9950 + 10050) or (101)50

Sol.
9950 + 10050 ↔ (101)50
10050 ↔ (1 + 100)50 – (1 – 100)50
10050 ↔ (C0 + C1x + C2x2 +…) – (C0 – C1x + C2x2…) (x = 100)
10050 ↔ 2[50C1 100 + 50C31003 +…+ 50C4910049]
10050 ↔ 2[50C1100 + 50C31003 +…] + 10050
Hence right hand side is bigger than left hand side
⇒ 9950 + 10050 < (101)50

Q. 4n
Show that : 2n–2Cn–2 + 2.2n–2Cn–1 + 2n–2Cn > , n∈N n > 2
n+ 1

Sol. Cn–2 ≥ 1, 2.2n–2Cn–1 ≥ 2, 2n–2Cn ≥ 1

2n–2

Adding, LHS ≥ 4
4n + 4 − 4 4
Binomial Theorem

Now, RHS = =4− <4

n+ 1 n+ 1
⇒ RHS < LHS.

17.
 9
3 1 
Q. If the term independent of x in the expansion of  x2 −
2
 is k, then 18k is
3x 
equal to
(A) 5 (B) 9 (C) 7 (D) 11

Ans. (C)

9−r r
3   −1 
Sol. Tr + 1 = 9Cr  x2 
2 
 
 3x 
9−r r r
3  1  1
= Cr   x 18−2r  −   
9

2  3 x
For term independent of x, 18 – 3r = 0 ⇒ r = 6
3 6
 3   −1  7 9
So, term independent of x = C6     =
2  3  18
7
∴ 18K = 18 × =7
18

Q. 22 23 24 2n+ 1 3n+ 1 − 1
Prove that: 2.C0 + .C1 + C2 + C3 + ... + Cn =
2 3 4 n+ 1 n+ 1

Sol. (M 1)
2 2

∫ (1 + x) ∫ (C )
n
dx = 0
+ C1x + C2 x2 + ... + Cnxn dx
0 0
2  
( )
n+ 1
1+ x  C 1x 2 x3 xn+ 1 
= C0 x + + C2 + ... + Cn 
n+ 1  2 3 n + 1

0

3 −1n+ 1
C .22 23
= 2.C0 + 1 + C2 . + ....
n+ 1 2 3

(M 2)
2r + 1 nCr
Tr =
Binomial Theorem

r+1
n n
2r + 1 nCr 2r + 1 n+ 1Cr + 1
S= ∑
r =0
r+1
= ∑ (n + 1)
r =0

18.
 n
1
= ∑
n + 1 r =0
 n+ 1 Cr + 1 2r + 1

n
1
= ∑
n + 1 r= 0
 n+ 1 Cr + 1 2r + 1

1  n+ 1
= 1 + C1 21 + n+ 1C2 22 + n+ 1C3 23 + ... + n+ 1Cn+ 1 2n+ 1 − 1
n+ 1 
1  
( )
n+ 1
= 1+ 2 − 1
n + 1  
3n+ 1 − 1
=
n+ 1

Q. C0 C1 C2 C3 Cn 1 + n.2n+ 1
Prove : + + + + ... + =
2 3 4 5 n+2 (n + 1)(n + 2)

Sol. (1 + x)n = C0 + C1x + C2x2 + …

1 1

∫x ( 1 + x ) ∫ (C x + C x )
n 2
dx = 0 1
+ C2 x3 + ... dx
0 0

Put (1 + x) = t
dx = dt
2 1
 x2 x3 x4 
∫( ) n
t − 1 t dt =  C0 + C1 + C2 + ... 
 2 3 4 
1  0
2 2
C C C C 
∫
1 1
∫
tn+ 1dt − tndt =  0 + 1 + 2 + ... n 
 2
 3 4 n + 2 
2 2
 tn + 2   tn + 1  C C C C
  −  = 0 + 1 + 2 + ... + n
 n + 2  1  n + 1  1 2 3 4 n+2

C C C

1  n+ 2
n+2 
2 − 1 −

1
n+ 1
2n+ 1 − 1 = 0 + 1 + 2 + ...
2 3 4
( )
C0 C1 C2  2 1  1 1
∴ + + + ... = 2n+ 1  − − +
2 3 4 n + 2 n + 1 n + 2 n + 1
Binomial Theorem

= 2n+ 1
( 2n + 2 − n − 2) + (n + 2 − n − 1)
(n + 2)(n + 1) (n + 1)(n + 2)
2n+ 1.n + 1
=
(n + 1)(n + 2)
19.
 Q. n

∑r .
2 n
Cr =
r= 0

Sol. (M 1)
n n n
∑
r =0
r.( r.nCr ) = n ∑r =1
r.n− 1 Cr − 1 = n ∑ (r − 1)
r =1
n− 1
Cr − 1 + n− 1Cr − 1

n n
= n(n − 1) ∑
r =2
n− 2
Cr −2 + n ∑
r=1
n− 1
Cr − 1

= n(n–1) 2n–2 + n.2n–1

= n.2n–2[(n – 1) + 2] = n.2n–2 (n + 1)
= n(n + 1)2n–2
(M 2)

(1 + x)n = C0 + C1x + C2x2 + C3x3 +…
Differentiating
n(1 + x)n–1 = C1 + 2C2x + 3C3x2 +…
nx(1 + x)n–1 = C1x + 2C2x2 + 3C3x3 +…
Differentiating
n[(1 + x)n–1 + x.(n – 1) (x + 1)n–2] = C1 + 22 C2. x + 32 C3x2 +…
put x = 1
n[2n–1 + (n – 1) 2n–2] = 12C1 + 22.C2 + 32 . C3 + …
n
∴ ∑r . C
r= 0
2 n
r
= n.2n–2 (2 + n – 1) = n(n + 1)2n–2

Q. Show that: S = C0 + 3.C1 + 5.C2 +…+ (2n + 1)Cn = (n+1).2n

Sol. S = 1. C0 + 3.C1 + 5.C2 +…+ (2n + 1) Cn

+ S = (2n + 1) Cn + (2n – 1) Cn–1 +…+ 1.C0
2S = (2n + 2) [C0 + C1 +…+ Cn]
2(n + 1)
Binomial Theorem

S= × 2n
2
∴ S = (n + 1)2n.

20.
 Q.  q + 1  q + 1  q + 1
2 n

Let Sn = 1 + q + q +…+ q and Tn = 1 +  2

+
n
 +…+   where q is a
 2   2   2 
real number and q ≠ 1.
If 101C1 + 101C2. S1 +…+ 101C101 . S100 = α.T100 then α is
(A) 299 (B) 202 (C) 200 (D) 2100

Ans. (D)

n+ 1
 q + 1
  −1
n+ 1
q −1  2  (q + 1)n+ 1 − 2n+ 1
Sol. Sn =
q− 1
; Tn =
q+ 1
=
(q − 1)2n
−1
2
101 q2 − 1 101 q3 − 1  q101 − 1 
Now C1 + 101C2 + C3 + ... + 101C101  
q−1 q− 1  q− 1 
 

=
1 
q − 1 
( 101
C1q1 + 101C2q2 + ... + 101C101q101 − ) ( 101
)
C1 + 101C2 + ... + 101 C101 


=
1
(q − 1)
{
[(1 + q)101 − 1] − [2101 − 1] = α.T100 }
(1 + q)101 − 2101 (q + 1)101 − 2101
⇒ =α
(q − 1) (q − 1)2100
α = 2100

Q.
n n

∑ k ∑ n
C k ·k2 n n
Ck
n
k =0 k =0
n
= 0 holds for some positive integer n then find ∑k + 1.
∑ k. C ∑
k =0
n
k
k =0
n
Ck .3k k =0

n(n + 1)
n(n + 1)2n−2
Sol. 2 =0
n.2n− 1 (1 + 3)n
Binomial Theorem

1
2n−2
n(n+1) 2 =0
n− 1 n n
n.2 2 .2

21.
 1 2n− 1
=0
n 2n+ 1
2n+1 – n.2n–1 = 0 ⇒ (4 – n) 2n–1 = 0
n=4
4 4
Ck 4 4
C1 4
C2 4 C 3 4 C4
∴
k =0
k+1 ∑= C0 +
2
+
3
+
4
+
5
1
=1+2+2+1+ = 6.2
5

Q.  C + C1   C1 + C2   C2 + C3   Cn− 1 + Cn  (n + 1)n
Prove :  0      ...   =

 C0   C1   C2   Cn− 1  n!

n− 1 n− 1 n+ 1
 Cr + Cr + 1  Cr + 1
Sol. LHS : ∏
r =0

 Cr
 =

∏
r =0
n
Cr
n− 1
n+ 1 1 (n + 1)n

= ∏
r= 0
r+1
= (n + 1)n ×
1.2.3.....n
=
n!

Q. Coefficient of x50 in (1 + x)101 (1 – x + x2)100 is-

Sol. (1 + x) (1 + x)100 (1 – x + x2)100 = (1 + x) [1 + x3]100

= coeff. of x50 in 1. (1 + x3)100 + coeff. of x50 in x(1 + x3)100
= 0 + 0 =0

Q. If α and β be the coefficients of x4 and x2 respectively in the expansion of

(x + x2 − 1)6 + (x − x2 − 1)6 then:
(A) α + β = 0 (B) α – β = – 132 (C) α + β = 60 (D) α – β = 60

Ans. (B)

Sol. (x + y)6 + (x – y)6 = 2{6C0x6 + 6C2x4y2 + 6C4x2y4 + 6C6y6}

∴ (x + x2 − 1)6 + (x − x2 − 1)6
Binomial Theorem

2{x6 + 6C2x4(x2 – 1) + 6C4x2(x2 – 1)2 + 6C6 (x2 – 1)3}

∴ α = 2[–6C2 – 2 6C4 – 3] = – 96
β = 2[6C4 + 3] = 36
∴ α – β = – 96 – 36 = – 132

22.
 Q. (1 + x + x2)n = a0 + a1x + a2x2 + … + a2nx2n
(i) Show that a0 + a1 + a2 + … + a2n = 3n

Sol. (M 1)
[1 + (x + x2)]n = nC0 + nC1(x + x2) + nC2(x + x2)2 +…
Put x = 1
(1 + 1 + 1)n = a0 + a1 +…+ a2n
∴ a0 + a1 +…+ a2n = 3n

(M 2)
[x2 + (x + 1)]n = nC0(x2)0(x + 1)n + nC1(x2)1(x + 1)n–1 + nC2(x2)2(x + 1)n–2 +…
put x = 1
(1 + 1 + 1)n = a0 + a1 + a2 +…
∴ a0 + a1 + a2 +…= 3n

(ii) a0 – a1 + a2 – a3 +…+ a2n = 1

Sol.

Put x = – 1 in given expression
[1 – 1 + (–1)2]n = a0 – a1 + a2 +…+ a2n
∴ a0 – a1 + a2 – a3+… = 1

(iii) a1 + a3 + a5 +…+a2n–1 =

Sol. (i) – (iii) gives

2(a + a +…) = 3 n
-1
1 3
3n − 1
∴ a1 + a3 + a5 +… =
2

3n + 1
(iv) a0 + a2 + a4 +…+ a2n =
2

Sol. (i) + (ii) gives

2[a + a +…] = 3 +1 n
0 2
3n + 1
a0 + a2 + a4 +… =
2
Binomial Theorem

23.
 (v) Show that ar = a2n–r ∀ r ∈ W

1
Sol. Replace x by x
n
 1 1  a 1 a2 a2n
 1 + + 2  = a0 + + 2 + ... + 2n
 x x  x x x
Multiply both sides by x2n
(1 + x + x2)n = a0x2n + a1x2n–1 + a2x2n–2 +…+ a2n
Also, (1 + x + x2)n = a0x0 + a1x1 + a2x2 +…+ a2nx2n …(A)
comparing coefficient of xr both sides ar = a2n–r

(vi) a 02 - a 21 + a 22 - a 23 + ... + a 22n = a n

1
Sol. Replaced x by –
x
n
 1 1  a 1 a2
 1 − + 2  = a0 − + ...
 x x  x x2
(x2 – x + 1)n = a0x2n – a1x2n–1 + a2x2n–2… …(B)
Multiply (A) and (B)
(1 + x + x2)n (x2 – x + 1)n = (a0 + a1x + a2x2 +…) (a0x2n – a1x2n–1+ a2x2n–2 –…)
(x4 + x2 + 1)n = a0{a0x2n – a1x2n–1 +…} + a1x{a0x2n – a1x2n–1 + a2x2n–2 +…} +
a2x2{a0x2n – a1x2n–1 + a2x2n–2 +…} +… …(C)
Taking coefficient of x from RHS-
2n

a02 − a21 + a22 − a32 + ... + a2n

2
…(D)
Replace x by x2 in (A)
(1 + x2 + x4)n = a0 + a1x2 + a2x4 +…+ anx2n +…+ a2nx4n …(E)
coefficient of x2n in RHS is an …(F)
from (F) and (D), a02 − a21 + a22 − a32 + ... + a22n = an .

(vii) a0a1 – a1a2 + a2a3 …. = 0

Sol. Taking coefficient of x from RHS in (C) & from RHS in (E)
2n–1
Binomial Theorem

–a a + a a – a a + … = 0
0 1 1 2 2 3
a0a1 – a1a2 + a2a3 – … = 0

24.
 (viii) a0a2 – a1a3 + a2a4 – … =

Sol. Taking coefficient of x

a a – a a + a a – … = a
2n–2
from RHS in (C) and (E)
0 2 1 3 2 4 n–1

Note : an–1 = an + 1 (∵ ar = a2n–r)

Q. If (1 + x)n = C0 + C1x + C2x2 +…+ Cnxn, then show that the sum of the products of
the Ci’s taken two at a time, represented by CiC j is equal to ∑∑
0≤ i < j ≤ n
2n!
22n− 1 −
2(n!)2

Sol.

To find : C0(C1 + C2 +…+ Cn) +
C1(C2 + C3 +…+ Cn) +
i=0
i=1
C2(C3 +…+ Cn) + i=2
 
 
 
 
Cn–1Cn i=n–1
n
(C0 + C1 + C2 +….+ Cn)2 = ∑C
r =0
2
r ∑ ∑ CC
+2
0≤i < j≤n
i j

(2n )2 − 2nCn

2
= ∑∑CC i j
0 ≤i< j≤n

(2n) !
∴ ∑∑CC i j
= 22n− 1 −
2(n!)2
0 ≤i< j≤n

n
Q. For any positive integers m, n (with n ≥ m) let   = nC m . Prove that
m
 
 n   n − 1  n − 2  m  n + 1 
  +   +   + ... +   =  
m  m   m   m   m + 1

Sol. (M 1)
Binomial Theorem

C + C
m

=( C
m
m+1

m+1
+…+ nCm
m
+ m+1
Cm) + m+2Cm +…+ nCm [∵ nCr + nCr+1 = n+1
Cr+1]
m+1
= ( Cm+1 +
m+2 m+2
Cm) + m+3Cm +…+ nCm
= ( Cm+1 +
m+3 m+3
Cm) +…+ nCm
n+1
m+1
= C
25.
 (M 2)
Coefficient of xm in(1 + x)m + (1 + x)m+1 + (1+x)m+2+…+ (1 + x)n
[(1 + x)n−m+ 1 − 1]
Coefficient of xm in (1 + x)m
(1 + x) − 1
Coefficient of xm+1 in (1 + x)m+n–m+1 = n+1
Cm+1

Q. Prove that
 n  n  n  n − 1 n n − 2 n n − k  n
2k     − 2k − 1    + 2k −2    ......( −1)k    =
 0  k   1   k − 1   2   k − 2   k   0   k 
             

Sol. LHS = ∑ (−1) 2

r= 0
r k −r n
Cr n−r Ck −r
  

↓
n! (n − r) ! k!
×
r !(n − r) ! (k − r) !(n − k) ! k !
n! k!
×
(n − k) ! k ! r !(k − r) !

= nCk kCr
k
LHS = n Ck ∑ (−1)
r= 0
r k
Cr 2k −r = nCk (2 − 1)k = nCk

Q. n
Show that ∑
k =0
n
Ck sinkx cos(n – k)x = 2n–1 sinnx

Sol.

Let s = C0 sin0 cosnx + C1 sinx cos(n–1)x + … + nCnsin(nx)cos(ox)
s = Cnsin(nx)cos(ox) + Cn–1sin(n – x)x cosx + … + C0 sin0 cos nx
+

2s = C0 (sin nx) + C1(sinnx) +…+ Cnsin(nx)

Point to Remember!!!
(C0 + C1 + ... + Cn ) 2n
s= sinnx = . sinnx
2 2 R=I+F
  = 2n–1 (sin nx). where
R = Real numbers,
Binomial Theorem

I = Integral part,
f = Fractional
part where f ∈ [0, 1)

26.
 Q. Find the integral part of (2 + 3)6

Sol. Let (2 + 3)6 = I + f, f > 0

+ & (2 − 3)6 = f1 0 < f1 < 1

(2 + 3)6 + (2 − 3)6 = I + f + f1
2[6C026 + 6C22431 + 6C42232 + 6C6 33] = I + (f + f1)
2[64 + 15 × 48 + 15 × 36 + 27] = I + (f + f1)
Even integer = I + (f + f1)
f1 + f = integer …(i)
(odd)
Now,
0<f<1
0 < f1 < 1
…(ii)
0 < f + f1 < 2
from (i) and (ii)
f + f1 = 1
∴ 2[64 + 720 + 540 + 27] = I + 1
I = 2701

Q. Find the value of (2 + 3)6 (1 − f), where f is fraction part of (2 + 3)6

Sol. From last question: 1 – f = f1 = (2 − 3)6

∴ (2 + 3)6 (2 − 3)6 = (4 − 3)6 = 1

Q. Prove that integer part of (3 + 7 )n is odd integer

(3 + 7 )n = I + f , 0 < f < 1 
Sol.  ⇒ 0 < f1 + f < 2 …(i)
+(3 − 7 )n = f1 , 0 < f1 < 1

2[nC03n + nC23n–271 + nC4 3n–472+…] = I + f + f1

Even integer = I + f + f1
Binomial Theorem

⇒ f + f1 = integer …(ii)
from (i) & (ii) f + f1 = 1 …(iii)
∴ even integer – (f + f1) = I
even integer – 1 = I
∴ I is an odd integer

27.
 Q. Show that integral part of (5 + 3 3)2n+ 1 is even.

Sol. (5 + 3 3)2n+ 1 = I + f
+ (5 − 3 3)2n+ 1 = – f1 f1 ∈ (0, 1)
2[2n+1C0 52n+1 + 2n+1
C2 52n–1 (3 3)2 +…] = I + f – f1
Even integer = I + f – f1
∴ f – f1 integer between – 1 & 1 ⇒ f – f1 = 0
 0<f<1 
 −1 < −f < 0 
 1 
 −1 < f − f1 < 1
 
 
∴ even integer = I + 0
∴ I = Even integer.

Q. Let I denotes the integral part & f the proper fraction part of (3 + 5)n where
n∈N and if ρ denotes the rational part and σ the irrational part of the same,
1 1
show that ρ = (I + 1) and σ = (I + 2f – 1)
2 2

Sol. (3 + 5)n = I + f = nC03n + nC13n–1 ( 5)1 +…

(3 − 5)n = f1 = nC03n – nC13n–1 ( 5)1 +…
∴ I + f = ρ + σ …(i)
f1 = ρ – σ …(ii)

(i) + (ii) given,

I + f + f1 = 2ρ
I + 1 = 2ρ (∵ f + f1 = 1)
I+ 1
ρ=
2
(i) – (ii) gives,
I + f – f1 = 2σ
I + f − (1 − f)
=σ
2
Binomial Theorem

I + 2f − 1
∴ σ =
2

28.
 Q. Sum of last 10 coefficients in the expansion of (1 + x)19

Sol.

+
S = 19C19 + 19C18 +…+ 19C10
S = 19C0 + 19C1 +…+ 19C9 (nCr = n
Cn–r)

2S = 219 ⇒ S = 218

Q. Sum of last 9 coefficients in the expansion of (1 + x)18

Sol.

+
S = 18C18 + 18C17 +…+ 18C10
S = 18C0 + 18C1 +…+ 18C8

2S= (18C0 + 18C1 +…+ 18C8 + 18C9 + 18C10 +…+ 18C18) – 18C9
218 − 18C9
  S=
2

Q. The remainder when (1523 + 2323) is divided by 19 is-

(A) 4 (B) 15 (C) 0 (D) 18

Ans. (C)

Sol. (19 – 4)23 + (19 + 4)23

= 2[23C01923 +23C2 1921 42 +…+ 23
c22 191422]
= a multiple of 19.
Hence, remainder = 0

1 10
Q. Find S =
2
C0 – 10C1 + 10C2 . 2 – 10C3 . 22 +…10C10 29

1 10
Sol. S=
2
[ C0 – 10C1 21 + 10C222 +…+ 10C10 210]

1 1
S= [1 − 2]10 =
2 2
Binomial Theorem

29.
 Q. Prove that: nC0 + n+1C1 + n+2C2 +…+ n+rCr = n+r+1Cr

Sol. LHS = coefficient of xn in (1 + x)n + (1 + x)n+1 +…+ (1 + x)n+r

(1 + x)n+r + 1 − (1 + x)n
= coefficient of xn in
1+ x − 1
= coefficient of xn+1 in (1 + x)n+r+1
= n+r+1Cn+1 = n+r+1Cr

To find Greatest Term

Q. 1
If x = , find the greatest term in the expansion of (1 + 4x)8
3

Sol. Let Tr+1 be greatest

Tr+1 ≥ Tr
Cr(4x)r ≥ 8Cr–1(4x)r–1
8

8! 8!
(4x) ≥
r !(8 − r) ! (r − 1) !(9 − r) !
41 1
≥ ⇒ 36 – 4r ≥ 3r
3 r 9−r
36
∴r≤ = 5.1 (approx)
7
at r = 5 T6 ≥ T5 Also, T7 ≤ T6
at r = 4 T5 ≥ T4 T8 ≤ T7
at r = 3 T4 ≥ T3 T9 ≤ T8
at r = 2 T3 ≥ T2
eat r = 1 T2 ≥ T1
Hence, greatest is T6
5
  1  8 × 7 × 6 45 57344
T6 = T5+1 = 8 C5 4    = × 5 =
3
    6 3 243
Binomial Theorem

30.
 Q. Find which term is numerically greatest term in (3 – 2x)9 at x = 1

Sol. |Tr+1| ≥ |Tr|

|9Cr39–r(–2x)r| ≥ |9Cr–1 310–r(–2x)r–1|
9! 9!
(2) ≥ 3
r !(9 − r) ! (10 − r) !(r − 1) !
Solving r≤4
∴ at r = 4 T5 ≥ T4 Also, T6 ≤ T5
at r = 3 T4 ≥ T3 T7 ≤ T6

at r = 2 T3 ≥ T2 
at r = 1 T2 ≥ T1 T10 ≤ T9
∴ T4 = T5 are greatest term.

Q.  3x 
10

Given T4 in the expansion of  2 +  has maximum numerical value. Find

 8 

range of x.

Sol. |T4| > |T5| and |T4| > | T3|

3 4 3 2
10 7 3x  10 6  3x  10 7 3x  10 8 3x 
C3 2   > C4 2   & C3 2   > C2 2  
 8   8   8   8 

1  3x  2 3x 1 2
  < × >
4 8  7 8 3 8
64
| x |< |x| > 2
21
 64   64 
x ∈ − , −2  ∪  2, 
 21   21 
Binomial Theorem

31.
 Binomial theorem for any index :
When n is a negative integer or a fraction then
the expansion of a binomial is possible only
when
(i) Its first term is 1, and
(ii) Its second term is numerically less than 1.
Thus when n ∉ N and |x| < 1, then it states
n(n − 1) 2 n(n − 1)(n − 2) 3
(1 + x)n = 1 + nx + x + x + …
2! 3!

n(n − 1)(n − r + 1) r
+ x + ...∞ ∞
r!
General Terms :
n(n − 1)(n − 2)...(n − r + 1) r
Tr + 1 = ⋅x
r!
Note :
(i) In this expansion the coefficient of different
terms can not be expressed as nC0, nC1, nC2, ...
because n is not a positive integer.
(ii) In this case there are infinite terms in the ex-
pansion.
(ii) In this case there are infinite terms in the ex-
pansion.

Some Important Expansions :

If |x| < 1 and n ∈ Q but n ∉ N, then
n(n − 1) 2 n(n − 1) … (n − r + 1) r
(a) (1 + x)n = 1 + nx + x +…+ x +…
2! r!
n(n − 1) 2 n(n − 1)(n − 2) 3 n(n − 1) … (n − r + 1)
(b) (1 − x)n = 1 − nx + x − x +…+ ( −x)r + …
2! 3! r!
n(n + 1) 2 n(n + 1)(n + 2) 3 n(n + 1) … (n + r − 1) r
(c) (1 − x)−n = 1 + nx + x + x +…+ x + ...
2! 3! r!
Binomial Theorem

−n n(n + 1) 2 n(n + 1)(n + 2) 3 n(n + 1) … (n + r − 1)

(d) (1 + x) = 1 − nx + x − x +…+ ( −x)r + ...
2! 3! r!

32.
 By putting mathrm n = 1, 2, 3 in the above results (c) and (d),
we get the following results

(e) (1 – x)–1 = 1 + x + x2 + x3 + ... + xr + ...

General term : Tr+1 = xr

(f) (1 + x)–1 = 1 – x + x2 – x3 + ... + (–x)r + ...

General term : Tr+1 = (–x)r

(g) (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + ... + (r + 1)xr + ...

General term : Tr+1 = (r + 1)xr

(h) (1 + x)–2 = 1 – 2x + 3x2 – 4x3 + ... + (r + 1)(–x)r + ...

General term : Tr+1 = (–x)r

(r + 1)(r + 2) r
(i) (1 − x)−3 = 1 + 3x + 6x2 + 10x3 + … + x +…
2!
(r + 1)(r + 2) r
General term = x
2!
(r + 1)(r + 2)
(j) (1 + x)−3 = 1 − 3x + 6x2 − 10x3 + … + ( −x)r + …
2!
(r + 1)(r + 2)
General term = ( −x)r
2!

1/ 2
 3 
Q. If |x| < 2 / 3 then the fourth term in the expansion of  1 + x
 2 
is
27 3 27 3 81 3 81 3
(A) x (B) − x (C) 256 x (D) − x
128 128 256

Sol. (A)
1 1 
1 / 2  − 1  − 2
2  2   3x 
3
27 3
T4 = ⋅  = x
3!  2 128
Binomial Theorem

33.
 2
 1 − x
Q. The term independent of x in the expansion of 
 1 + x 
is

(A) 4 (B) 3 (C) 2 (D) 1

Sol. (D)

(1 – x)2 (1 + x)–2

⇒ (1 – 2x + x2) (1 – 2x + x2 + ...)

⇒ so term independent of x = 1.

Q. The coefficient of x5 in the expansion of (1 – x)–6 is

(A) 1260 (B) – 1260 (C) – 252 (D) 252

Sol. (D)

x5 occurs in T6 of the expansion, so

6.7.8.9.10 5
x = 252x5
T6 = T5 + 1 =
5!

∴ Coefficient of x5 = 252

Application of binomial theorem :

(i) With the help of binomial theorem, we
can find out the value of sq. root, cube
root and 4th root etc. of the given number
upto any decimal places.

Q. The value of cube root of 1001 upto five decimal places is

(A) 10.03333 (B) 10.00333 (C) 10.00033 (D) None of these

Sol. (B)  11

−

1

1/ 3  
 1   1 1 3  3  1 
(1001)1/ 3 = (1000 + 1)1/ 3 = 10  1 + = 10 1 + ⋅ + + …
 1000   3 1000 2! 10002

 
Binomial Theorem

 
= 10 {1 + 0.0003333 – 0.00000011 + ...}
= 10.0033

34.
 (ii) To find the sum of Infinite series :

We can compare the givein infinite series with

the expansion of (1 + x)n = 1 + nx +

n n−1 2
x
( )
2!
+ ... and by finding the value of x and n and

putting in (1 + x)n the sum of series is deter-

mined.

1 1.3 1.3.5
Q. The sum of 1 +
+ +
4 4.8 4.8.12
+ … ∞ is

1
(A) 2 (B) (C) 3 (D) 23/2
2

Sol. (A)
n(n − 1) 2
Comparing with 1 + nx +
2!
x +…

nx = 1/4 ...(i)
n(n − 1)x2
and
2!
= 1.3 / 4.8

nx(nx − x) 3 11  3
or
2!
=
32
⇒  − x =
4 4  16
(by (i))

1  3 1 3 1
⇒  4 − x = 4 ⇒ x = 4 − 4 = − 2 ...(ii)

Putting the value of x in (i)

n(–1/2) = 1 / 4 ⇒ n = – 1/2

∴ sum of series = (1 + x)n

−1/ 2
 1
=
 1 − 2  (
= 1/2 )−1/ 2 = 2
Binomial Theorem

35.
 (iii) Approximation :

Q. If x is so small so that its square and higher power can be neglected. Find the
−5
 2x 
 1 + 3  + (4 + 2x)1/ 2
value of .
(4 + x)3/ 2

Sol. (D)
−5
 2x   10x   x
 1 + 3  + (4 + 2x)1/ 2  1 − 3  + 2  1 + 4  −1
= =
1
3−
10x x 
+ 
 3x 
 1 + 8 
(4 + x)3/ 2  3x  8  3 2
81 +
 8 
3 17x   3x  3  17x 3x  72 − 95x
=  1−   1−  = 1 − − =
8 18   8 8 18 8  24 × 8
Binomial Theorem

36.