MATHEMATICAL INDUCTION & BINOMIAL

FIRST PRINCIPLE OF FINITE INDUCTION:

Suppose that to each n  N there corresponds a proposition (or theorem or formula)
P(n) which is either true or false.
If (i) P(1) is true and
(ii) P (k) is true  P(k + 1) is true  k  N, then P(n) is true for  k  N.

Working Method:
A proof of P(n) by the method of induction involves the following two steps:
Step-1: Show that P(1) is true.
Step-2: Assuming that P(k) is true for some k  N, show that P(k + 1) is true.

Then from the step 1 and 2, it follows that the proposition P(n) holds good  n  N.

ILLUSTRATIONS:
n ( n  1)( 2n  1)
(1) n² = nN
6
(2) 102n1 + 1 is divisible by 11
(3) 72n  42n is divisible by 33
(4) 11n+6 + 122n+9 is divisible by 133
Proof: P(1) = 117 + 1211
= 112 · 115 + 122 · 129 [133 = 112 + 12 = 122 – 11]
= 115 [133 – 12] + 129[133 + 11]
= 133(115 + 129) + 12 · 11[114 – 128]
= 133(115 + 129) –12 · 11[128 – 114]

(5) xn  yn is divisible by x  y where n is a positive integer & is divisible by

x + y if n is even natural number .
n
(6) 22 + 1 ends in the digit 7  n  2 ; n  N .
n
(7)  nCr = 2n for n  1 ; n  N
r 0

(8) If x is not an integral multiple of 2, use mathematical induction to prove that
cos x + cos 2 x + ..... + cos n x = cos n 2 1 x sin nx2 cosec x2 . [JEE’94, 4]
(9) For every natural number n , prove by mathematical induction that
4n + 15 n  1 is divisible by 9 . [REE’94, 6]

Home Work:
n ( n  1)(2n  7)
(1) 1.3 + 2.4 + 3.5 + ...... + n (n + 2) = where n  N.
6
1 1 1 1 n
(2)    .........   where n  N.
2.5 5.8 811
. (3n  1)(3n  2) 6n  4
(3) 3n
2  1 is divisible by 7
(4) 32n + 7 is divisible by 8
(5) n3 + (n + 1)3 + (n + 2)3 is divisible by 9
(6) 5n+3 + 113n+1 is divisible by 17
(7) 52n+1 + 3n+2 2n1 is divisible by 19
(8) 6n+2 + 72n+1 is divisible by 43

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 1 1 1 tan (n  1)   tan 
(9)  
cos  cos3 cos  cos5 cos  cos7 
+.... upto n terms = 2 . sin 
 nN

10 n 1  9n  10
(10) 3 + 33 + 333 + ..... + (333.....n times) =  n  N.
27
[Sol. Let P(n) be
10 n 1  9n  10
3 + 33 + 333 +....to n terms = ....(A)
27
Step-I : Let n = 1
Then L.H.S. of (A) = 3 and
10 2  9  10 81
R.H.S. of (A) = = =3
27 27
 (A) is true for n = 1.
Step-II: Suppose P(k) is true,
10 k 1  9k  10
i.e. 3 + 33 + 333 +....to k terms = ....(B)
27
Now 333 .... 3 (3 written k + 1 times)
= 3 × 10k + 3 × 10k – 1 + 3 × 10k – 2 +....+ 3 × 101 + 3 × 100
= 3(10k + 10k – 1 +....+ 101 + 1)
3(10 k 1  1) 3(10 k 1  1)
= = ....(C)
10  1 9
Add 33....3(k + 1 3's) to both sides of (B).
 3 + 33 +....+ 333...(k + 1 terms)
10 k 1  9k  10 3(10 k 1  1)
=  , using (C)
27 9

10 k 1  9k  10  9  10 k 1  9 10 k 1 (1  9)  9(k  1)  10 10 k  2  9(k  1)  10
= = = ....(D)
27 27 27
(D) shows that P(k + 1) is true if P(k) is true.
 Steps I and II show that P(n) is true  n  N. ]
n 4 n 3 11n 2 n
(11)    is a natural number  n  N.
24 4 24 4

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 TEACHING NOTES
SYLLABUS:
CBSE: Histroy, statement and proof of the binomial theorem for positive integral indices.
Pascal's triangle, general and middle term in binomial expansion, simple
applications.
AIEEE: Binomial theorem for a positive integral index, general term and middle term,
properties of Binomial coefficients and simple applications.
JEE 2009: Binomial theorem for a positive integral index, properties of binomial coefficients.
BINOMIAL
1 ST LECTURE
(1) Binomial Expression : An algebraic expression consisting of two different terms is
called a Binomial Expression.
1
e.g. (i) x + y (ii) x3 + x2 (iii) x2 +
x2
But x + 3x is not a binomial it is a monomial.
(2) Binomial Theorem : The formula by which any positive integral index (power) of a
binomial expression which can be expanded in the form of a series is known as Binomial
Theorem. (This theorem was given by Newton)

Historical Development:
(x + y)2 = (x + y)(x + y) = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Earlier people used to multiply the brackets to expand the given binomial of known
index. Then came the Pascal's triangle.
(x + y)5 = x5 + x4y + x3y2 + x2y3 + xy4 + y5

(3) Newton's Binomial Theorem Expansion:

(x + y)n = nC0xn + nC1xn – 1 · y + nC2xn – 2 · y2 +.....+ nCrxn – r · yr +.....+ nCn xn – n · yn
T1 = nC0 xn – 0 y0 ; T2 = nC1 xn – 1 y1 ........ etc

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  General term in the expansion of
n
(x + y)n is Tr + 1 = Cr · x n  r · y r  To be remembered.
 Abbreviated form of Binomial theorem is
n
(x + y)n =  n Cr · x n  r · yr
r0

n
Proof: (x + y)n =  n Cr x n r y r
r 0

P(1) is true (verify) k

Let P(k) be true i.e. (x + y)k =  k Cr x k r y r is true
r 0
k 1
To prove P(k + 1) is true i.e. (x + y)k + 1 =  k1C r x k1r y r
r 0

k 1
R.H.S. =  ( k Cr  k C r1) x k1r yr
r 0

 k 1   k 1 
x   k C r x k  r y r   y   k C r 1x k  ( r 1) y r 1 
   
 r 0   r 0 
x P(k)  + *
(for the 1st bracket kCk + 1 = 0 and for 2nd bracket kCr – 1 = 0 for r = 0)
* Let r–1=t
 k 
y   k C t x k  t y t  = y P(k ) 
 
 t 1 
 R.H.S. = (x + y)P(k) = (x + y)(x + y)k = (x + y)k + 1 = L.H.S. Hence proved.]
(4) Observations / General highlights of Binomial Theorem :
(i) Number of terms in the expansion of (x + y)n is (n + 1) i.e. one more than the index.
or we can also say it using beggar method n coins and 2 beggars.
 n + 1C  (n + 1) times.
1
Find the number of terms in the expansion of (x + y + z)8;
[Hint: 8 + 2C2 = 10C2 = 45 Ans. ]
(ii) Sum of the indices of 'x' and 'y' in each term in the expansion of (x + y)n is 'n'.
(iii) nC nC nC
..........nCn are called binomial coefficients or combinatorial coefficients
0, 1, 2,
and may be simply written as C0, C1, C2, ....... Cn.
(x + y)n = nC0 xn y0 + nC1 xn – 1 y + nC2 xn – 2 y2 + ...... + nCn xn – n yn.
Find the sum of all the combinatorial coefficient.
i.e. nC0 + nC1 + nC2 + ...... + nCn = 2n.
Put x = 1 and y = 1 to get sum of all the binomial coefficient.
In (x + 2y)2 find the sum of all the coefficients.
(x + 2y)2 = x2 + 4xy + 4y2
 sum of all coefficient = 1 + 4 + 4 = 9
we can also get it by putting x = y = 1
(1 + 2)2 = 9

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  sum of all binomial coefficients in
(x + y)n = 2n  In this case sum of coefficient = sum of binomial coefficient.
(iv) Binomial coefficients of the term equidistant from beginning and end are equal.
(v) coefficient of xr in the expansion of (1 + x)n is nCr and Tr + 1 = nCr xr
 coefficient of (r + 1)th term = coefficient of xr = nCr in the expansion of (1 + x)n.

(6) Important terms in Binomial expansion:

I. General Term: (Tr + 1)th term is called the general term.

Tr 1  n Cr · x n  r · y r
(A) Particular term :
7
 y
(i) Find the fourth term in the expansion of  2 x   .
 2
(ii) 6 2 3
Find the coefficient of x in (1 + 3x + 3x + x ) 15 [Ans. 45C6]
[Sol.
(i) Tr + 1 = nCr xn – r yr n=7
3
7C  y
T3 + 1 = 3 (2x)4    = – 35 × 2x4y3 = – 70x4y3 Ans.
 2
(ii)  Tr + 1 = 45Cr xr
 r=6  coefficient is 45C6 ]

(B) Term independent of x :

6
 2 1 
(i) Find the term involving x3 in  2 x   .
 3x 

 1 
r
(2)6 r
[Sol. Tr + 1 = 6Cr (2x2)6 – r    = 6Cr · · x12 – 2r – r
 3x  (3) r
 12 – 3r = 3;  3r = 9  r = 3
 T4 or the fourth term will have x3

(2)3 20 8 3 160 3

T4 = – 6C
3 ·
· x3 = – x =– x Ans. ]
33 27 27

12
 2 1
(ii) Find the value of the term is independent of x in the expansion of  x   .
 x
r
1
[Sol. Tr + 1 = 12Cr (x2)12 – r   = 12Cr x24 – 3r ;  24 – 3r = 0  r=8
x
 T9 is independent of x
T9 = 12C8 = 495 Ans. ]

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 1
(iii) Find the greatest value of the term independent of x in (x sin  + cos )10,   R
x
r 10
10C 1  C5
[Sol. Tr + 1 = r (x sin )10 – r  cos   = 10Cr x10 – 2r (sin )10 – r (cos )r [Ans. ]
x  25
10 – 2r = 0  r = 5 ;
T6 = 10C5 sin5 · cos5 =
10
1 10C 5
C5
T6 = 5 (sin 2) = (sin 2)5
25 2 5

10
C5
 maximum value = Ans. ]
25
(C) Middle Term :
11
 2 1 
(i) Find the middle term in the expansion of  2 x  
 3x 
[Sol. No. of terms = 12
 6th and 7th are the middle terms. Middle term has greatest binomial coefficient.]
7
 1
(ii) Find the middle term / terms in the expansion of  x   .
 x
[Sol. Number of terms is 8
 middle terms are 4th and 5th term
3 4
7C
1
 1 35
T4 = 3 x4  = 35x; T5 = 7C4 x3   = ]
x x x
2n
 1
(iii) Middle term in the expansion of  x   is
 x
( 2n )! 2 · 6 ·10......(4 n  2)
(A*) (B*)
n!· n! n!
n 1 n  2 nn 2n
(C*) · ...... (D*) [1 · 3 · 5 · ..... (2n – 1)]
1 2 n n!
[Sol. Number of terms = 2n + 1
 middle term is (n + 1)th term
1 ( 2n )!
 Tn + 1 = 2nCn xn · = 2nCn =  (A)
xn n!· n!
2n n !(1· 3..........(2n  1))
=  (D)
n ! n!
1· 2 ·3......n ( n  1)(n  2)...(n  n )
=  (C)
n!· n!
2 · 6......(4n  2)
=  (B) ]
n!

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 n
 1 
Q.18bin If the coefficients of the first three terms in the expansion of the binomial  x  4 
 2 x

are in A.P. then which of the following statement(s) is/are correct?

(A*) Number of terms in the expansion of the binomial is 9.
35
(B*) Coefficient of middle term is .
8
(C*) There is no term independent of x.
(D*) The number of terms in the expansion with integral powers of 'x' is 3.
n
 1 
[Sol.  x  1 4  [12th, 14-06-2009, P-1]
 2·x 
n 2
1 1
n
C0  x
n n
 C1 x   n 1
· 3 4  nC2
2x
 x 2 ·
4 x
 ......

n n
nC C1 C2
now 0, , are in A.P..
2 4
n n (n  1) n (n  1) n (n  1)
1, , ; n=1+  (n – 1) =  n=8
2 8 8 8
8 r
 1 2 x 1 4  8 r 
hence we have  x  2  ; Tr+1 = 8 C x 4
r
·x 2 · r
  2
8r r
for integral powers of x,  must be an integer
2 4
16  2r  r 16  3r
 must be an integer  r = 0, 4, 8
4 4
 3 terms with integral power  (D)]
(D) General Problems :
(i) Find the value of (1 + 7 7
2 ) + (1 – 2)
[Sol. (1 + 7 7 = 2[7C0 + 7C2 ( 2 )2 + 7C4 ( 2 )4 + 7C6 ( 2 )6 ]
2 ) + (1 – 2)
= 2[1 + 21 · 2 + 35 · 4 + 7 · 8]
= 2[1 + 42 + 140 + 56]
= 2[239] = 478 Ans. ]

(ii) Find the number of divisors of ( 3 + 3)5 – ( 3 – 3)5.

[Sol. ( 3 + 3)5 – ( 3 – 3)5 = 2[5C0 · 35 + 5C2 · 33 ( 3 )2 + 5C4 · 31( 3 )4]
= 2[243 + 10 · 81 + 5 · 27]
= 2[243 + 810 + 135] = 2 × 1188 = 2376 = 23 33 11  32 Ans.]
(iii) In the expansion of (1 + x)10, the coefficient of (4r + 5)th term is equal to the coefficient
of (2r + 1)th term find r.
[Sol. Tp + 1 = 10Cp xb
T4r + 5 = 10 C4r + 4 x4r + 4 ; T2r + 1 = 10 C2r x2r
 10C 10
4r + 4 = C2r
 4r + 4 = 2r 4r + 4 + 2r = 10
2r = – 4 (rejected) 6r = 6  r = 1 ]
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(iv) If the ratio of 7th term from beginning and 7th term from the end in the expansion of
1
(21 3  31 3 )n is equal to then the value of n is equal to
6
(A*) 9 (B) 11 (C) 7 (D) 10
n 6
[Sol. T7 from beginning = nC6 (21/3)n – 6 (3–1/3)6 = nC6 · 2 3 · 3–2
6 n
T7 from end = nC6 · 3 3 · 22.
n 6
n n  6  6 6 6  n
1 C6 · 2 3 ·3 2 1
Ratio = ;  6n
= ; 2 3 ·3 3 = 2–1 · 3–1
6 6
n
C6 · 3 3 · 22

n  12  12  n
=–1 and =–1
3 3
n – 12 = – 3  n = 9 n = – 3 + 12 = 9
 n = 9 Ans. ] n
 1
(v) If the sum of all the coefficients in the expansion of  x   is 4096 then find the
 x
greatest binomial coefficient.
[Sol. 2n = 4096;  n = 12
 No. of terms = 13
 Middle term is 7th and middle term has greatest binomial coefficient i.e. 12C6.]
20
 13 1 
(vi) In the expansion of  4  1 4  which of the following hold good?
 6 
(A*) number of irrational term = 19 (B*) middle term is irrational
(C*) number of rational term = 2 (D) none
r 40  2 r r r 160 11r  r
20 1 3 20  r  1  20
 
[Sol. Tr + 1 = C r ·( 4 )  14 = Cr ·2 3 ·2 4 ·3 4 = 20
Cr ·2 12 ·3 4
6 
160  11r r
for rational terms and – must be integer, r  [0, 20]
12 4
r
for to be an integer r Î {0, 4, 8, 12, 16, 20}
4
160  11r
for to be an integer r  {8, 20}
12
 common solution is r = 8 or r = 20
 only 9th and 21st terms are rational.
 remaining 19th term are irrational and middle term is 11th which is irrational
 A, B, C are all correct. ]
(vii) In the expansion of y = 1 + (1 + x) + (1 + x)2 + ....... (1 + x)19.
if the coefficient of xp is the greatest coefficient then find the value of p (p  N).
(1  x )20  1 (1  x )20  1
[Sol. y = =
1  x 1 x
Now coefficient of xp in y is same as coefficient of xp+1 in (1 + x)20 – 1
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 which is 20Cp+1 and it is greatest
 p + 1 = 10  p = 9 Ans. ]
(viii)101/bin If the expansion, (3x8 – 2x6 + x5 + 2x4 – x2 + 1)5  a0 + a1x + a2x2 + ......+a40x40,
20
then the sum  a 2r equals
r 0

(A) 1056 (B*) 528 (C) 508 (D) 1058

[Sol. Put x = 1 a0 + a1 + a2 + ...... + a40 = 45 = 1024 [11th, 25-01-2009, P-2]
Put x = –1 a0 – a1 + a2 + ...... + a40 = 25 = 32
20
2(a0 + a2 + ....... + a40) = 1056;  a 2r = 528 ]
r 0
Note: an + bn is divisible by (a + b) only if n is odd.
e.g. 1125 + 1225 when divided by 23 leaves the remainder zero.
(E) NUMERICALLY GREATEST TERM:
EXAMPLES:
1
(1) If x = find the greatest term in the expansion of (1 + 4x)8.
3
[Sol. Replace x|x|
8
Tr 1 Cr (4 | x |) r (r  1)!(9  r )!| 4 x | 9r 4
Consider = 8 r 1 = = 
Tr Cr 1(4 | x |) r!(8  r )! r 3

Tr 1 36  4r
 Tr = 3r ; Tr + 1  Tr so log as 36 – 4r  3r; 36  7r

The value of r consistent with the inequality.

T2 > T1 ; T3 > T2 ; T4 > T3 ; T5 > T4 ; T6 > T5 ; T7 < T6
 T6 is numerically greatest terms.
5
8C  4
T6 = 5 (4x)5 = 8C
5   ]
 3

(2) If x = 1 find numerically the greatest term in the expansion of (3 – 2x)9.

9
Tr 1 Cr (3)9 r (2x )r (r  1)!(10  r )!| 2 x | (10  r )2 20  2r
[Sol. T = 9 9  r 1 r 1 = = =
r C r 1 (3) (2x ) (9  r )! r! 3 3r 3r

Tr 1 
T according as 20 – 2r 3r
Tr  r
20  5r or 4  r
T2 > T1 ; T3 > T2 ; T4 > T3 ; T5 = T4 ;
T6 < T5
 T4 and T5 are numerically greatest terms. ]

Bansal Classes Page # 9

 10
 3x 
(3) If 4th term T4 in the expansion of  2   has maximum numerical value. Find the
 8 

 64   64 
range of 'x'. [Ans. x    ,2    2,  ]
 21   21 
2
 3x 
[Sol. Given (1) T4 > T3 and (2) T4 > T5 : Tr + 1 =10Cr 210 – r  
 8 

3 4
3 | x |  10 3| x | 4!· 6!· 2  3 | x |  4 · 2 3 | x |
Solving (2) 10C (2)7 
 6
 > C4(2)   ;  
3
 8   8  3!· 7! >  8  ; 7 > 8

64 64  64 64
> | x |;  |x|<  x
21 21 21 21

3 2
 3 | x |  10 3| x | 
Solving (1) 10C3 27   > C2 28  
 8   8 

2!·8! 3 | x |
·
3!· 7! 8 > 2  | x | > 2
x  (– , – 2)  (2, )
 64   64 
 (1)  (2) is x    ,  2    2,  ]
 21   21 
n
 x 2
(4) Find the index 'n' of the binomial    if the 9th term of the expansion has
 5 5
numerically the greatest coefficient (n  N). [Ans. n = 12]
n r 2
nC x 2
[Sol. (1) T9 > T10 & (2) T9 > T8 ; Tr + 1 =  
2 5
 
5

n 8 8 n 9 9
nC 
x  2 x  2
(1) 8 5
   > nC9    
  5 5 5
don't consider x.
n! x n! 2 9
 ; put x = 1 to get > 2  9 > 2n – 16  n < 12 · 5
8 ! n  8 ! 5 9 ! n  9  ! 5 n 8
n 8 8 n 9 7 n 9 n 8
x  2 x  2 n  7 2 1
(2) nC  
8 5   > nC7     ; · > 
  5 5 5 8 5 5

2n  14 1
or
8 ·5 > 5 ; 2n – 14 > 8  2n > 22  n > 11

 n = 12 (n  N) ]

Bansal Classes Page # 10

(G) BINOMIAL COEFFICIENT:
If C0, C1, C2, ...... Cn are combinatorial coefficient in the expansion of (1 + x)n : n  N
then
(i) (1 + x)n = C0 + C1x + C2x2 + C3x3 + ........ + Cnxn
Put x = 1
 sum of all the binomial coefficient in the expansion of (1 + x)n is 2n.
C0 + C1 + C2 + ........ + Cn = 2n
(ii) Put x = – 1
C0 – C1 + C2 – C3 + ........ (–1)nCn = 0
 C0 + C2 + C4 + ....... = C1 + C3 + ......... = 2n – 1

EXAMPLES:
(2n)!
(1) Prove that C02  C12  C22  C32  .......  C2n = 2nCn = (To be remembered)
n!n!
[Sol. (1 + x)n = C0 + C1x + C2x2 + .................. + Cnxn
(x + 1)n = C0xn + C1xn–1 + C2xn–2 +........ + Cnx0
coefficient of xn in (1 + x)n × (1 + x)n will give C02  C12  C22  .......  C 2n
and coefficient of xn in (1 + x)2n is also equal to 2nCn
 C02  C12  C22  C32  .......  C2n = 2nC
n Hence proved ]

(2) Find the sum

S = C0C1 + C1C2 + C2C3 + ..........Cn–1Cn.
[Sol. (1 + x)n = C0 + C1x + C2x2 + .......... + Cnxn
(x + 1)n = C0xn + C1xn–1 + C2xn–2 +........ + Cnx0
————————————————————
(1 + x)2n = (C0 + C1x + C2x2 + .... + Cnxn) (C0xn + C1xn–1 + C2xn–2 +.....+ Cnx0)
now C0C1 + C1C2 + C2C3 + ..........Cn–1Cn appears in the coefficient of xn – 1 or
xn + 1 in (1 + x)2n
i.e. it is 2nCn–1 or 2nC
n+1

(2n )! (2n )!
i.e. =
(n  1)!(n  1)! (n  1)!(n  1)! ]

(3) Find the sum of the series (using Fredril Karl Gauss)
S = 1 · nC1 + 2 · nC2 + 3 · nC3 + ........... + (n – 1) nCn – 1 + n · nCn.
[Ans. n · 2n – 1]
[Sol. Method-I:
S = 0 · nC0 + 1 · nC1 + 2 · nC2 + ........... + n · nCn
S = n · nC0 + (n – 1) · nC1 + (n – 2) · nC2 + ........... + 0 · nCn
—————————————————————————
2S = n [ nC0 + nC1 + nC2 + ..........+ nCn ]
n n
 S= ·2  S = n · 2n  S = n · 2n – 1 Ans.
2

Bansal Classes Page # 11

 Method-II: Algebraic Method
Tr = r · nCr
n n r · n! n n · (n  1)! n
n
S =  r · Cr =  r!(n  r )! =  (r  1)!(n  r )! =  n 1Cr 1
n
r 1 r 1 r 1 r 1
S = n [ n–1C0 + n–1C1 + n–1C2 + ....... n–1Cn–1 ]
 S = n · 2n – 1 Ans.
Method-III: Calculus Method
(1 + x)n = nC0 + nC1x + nC2x2 + .......... + nCnxn
differentiation of an identity is also an identity
 n(1 + x)n – 1 = nC1 + 2 · nC2x + 3 · nC3x2 + .......... + n · nCnxn – 1
now put x=1
then C1 + 2C2 + 3C3 + .......... + nCn = n · 2n – 1 Ans. ]
(4) Find the sum (Fredril Karl)
S = 1C0 + 2C1 + 3C2 + 4C3 + ...... + (n + 1)Cn. [Ans. (n + 2)2n–1]
[Sol. S = 1C0 + 2C1 + 3C2 + ......+ (n + 1)Cn
S = (n + 1)C0 + nC1 + (n – 1)C2 +.............. + 1Cn
————————————————————
2S = (n + 2) [C0 + C1 + C2 + ....... + Cn]
S = (n + 2)2n – 1
OR consider
(1 + x)n = C0 + C1x + C2x2 + .......... + Cnxn
multiply both sides by x
(1 + x)n · x = C0 x + C1x2 + C2x3 + .......... + Cnxn + 1
and differentiate using product rule
[n(1 + x)n – 1 x + (1 + x)n] = C0 + 2C1x + 3C2x2 + .......... + (n + 1)Cnxn
put x = 1
n  n
C0 + 2C1 + 3C2 + .......... + (n + 1)Cn = n · 2n – 1 + 2n =   1 2 = (n + 2)2n–1 ]
2 
(5) Find the sum (Fredril Karl)
S = 1 nC0 + 3 nC1 + 5 nC2 + .......... + (2n + 1)nCn. [Ans. (n + 1)2n]
[Sol. S = 1 nC0 + 3 nC1 + 5 nC2 + .......... + (2n + 1)nCn
S = (2n + 1)nC0 + (2n – 1)nC1 + (2n – 3)nC2 + .................... + 1 nCn
————————————————————————————
2S = (2n + 2) [ nC0 + nC1 + nC2 + ...... + nCn ]
 S = (n + 1)2n Ans. ]
(6) Find the sum (Fredril Karl)
S = 1( nC0 )2 + 3( nC1 )2 + 5( nC2 )2 + .......... + (2n + 1)( nCn )2.
[Ans. (n + 1)2nCn]
[Sol. S = 1( nC0 )2 + 3( nC1 )2 + 5( nC2 )2 + ..... + (2n + 1)( nCn )2.
S = (2n + 1)( nC0 )2 + (2n – 1) ( nC1 )2 + (2n – 3)( nC2 )2 +...... + 1( nCn )2
———————————————————————————————
2S = (2n + 2)( nC0 )2 + (2n + 2)( nC1 )2 + .............................. + (2n + 2)( nCn )2 ]
S = (n + 1) [ C02  C12  ......  C 2n ]
S = (n + 1)2nCn Ans. ]
Bansal Classes Page # 12
N.C.E.R.T

 C0  C1   C1  C2   C 2  C3   C  Cn 
(7)  ×   ×  × .........×  n 1 
 C   C   C   C 
 0   1   2   n 1 

n nC n n n 1 C n
(n  1)(n  1  r )! n (n  1)
r 1  C r
[Sol. P=  = n r
= = 
r 1
n
Cr 1 r 1 C r 1 r 1 r (n  1  r )! r 1 r

1 1 1 1 ( n  1) n
 P = (n + 1)n     ......   = ]
1 2 3 n n!
n 1 n r
(8) If an =  n then find the value of n .
r 0 Cr r 0 Cr

0 1 2 n
[Sol. S = n
 n
 n
 ........  n
C0 C1 C2 Cn

n n 1 n  2 0
S= n
   ........ 
C0 n C1 n C2 n
Cn
———————————————
n n n n n ·an
2S = n
   ........  ; 2S = n · an  S= Ans. ]
C0 n C1 n C 2 n
Cn 2

(9) If (1 + x + x2)n = a0 + a1x + a2x2 + ........ + a2nx2n (n  N), then

(a) a0 + a1 + a2 + a3 + ........ + a2n = 3n put x = 1
(b) a0 – a1 + a2 – a3 + ........ + a2n = 1 put x = – 1

3n  1
(c) a1 + a3 + a5 + ........ + a2n – 1 = ((a)– (b)  c)
2

3n  1
(d) a0 + a2 + a4 + ........ + a2n = ((a)+ (b)  d)
2

(e) a 02 – a12 + a 22 – a 32 ........ + a 22n = an

[Proof: Replace x by – 1/x
n a 2n
 1 1 a1 a2
1    = a – + – ............. –
 x2 x 
0
x x x 2n
(1 + x2 – x)n = a0 x2n – a1 x2n – 1 + a2 x2n – 2 – ...... + a2n ....(1)
(1 + x + x2)n = a0 + a1x + a2x2 + ........ + a2nxn ....(2)
(1)×(2) —————————————————————————

in (1  x 2 )2  x 2 n coefficient of x 2n gives a 02 – a12 + a 22 ...... + a 22n

also [(1 + x2)2 – x2]n = [1 + x4 + 2x2 – x2]n = [1 + x2 + x4]n
= a0 + a1x2 + a2x4 + a3x6 + ....... + a2nx4n
 coefficient of x2n will be an
 a 02 – a12 + a 22 ........ + a 22n = an ]
Bansal Classes Page # 13
(f) a0a1 – a1a2 + a2a3 + ........ = 0
coefficient of x2n + 1 in the expansion of [1 + x2 + x4]n. But in its expansion we always
have x(even power).
 coefficient of x2n + 1 = 0

(g) a0a2 – a1a3 + a2a4 ........ = coefficient of x 2 n  2 x 2 n  2 in [1 + x2 + x4]n

which is an + 1 or an – 1.
n
(10) Prove that:  n Ck sin kx cos(n  k) x = 2n – 1 sin nx
k 0
[Sol. S = 0 · sin 0x · cos nx + nC1 · sin 1x · cos (n – 1)x + ..... + nCn · sin nx · cos 0x
nC

S = nC0 · sin nx · cos 0x + nC1 · sin (n – 1)x · cos 1x + ..... + nCn · sin 0x · cos nx
—————————————————————————————————
2S = nC0 sin (0x + nx) + nC1 · sin x  ( n  1) x  + ............ + nCn · sin nx  0 x 
2S = [nC0 + nC1 + nC2 + ...... + nCn] sin nx
S = sin nx · 2n – 1 ]
(11) Use binomial theorem to prove that 11n – 10n – 1 is divisible by 100  n  N.
[Sol. (10 + 1)n – 10n – 1
(1 + 10)n – 10n – 1
[1 + nC1 (10) + nC2 (10)2 + nC3 (10)3 + ........ + nCn (10)n ] – 10n – 1

100[ n C2  n C3 ·10  ......  n Cn ·10 n  2 ]

 
integer

 always a number divisible by 100. ]

(12) 42n – 15n – 1 is divisible by 225  n  N
[Sol. (1 + 15)n – 15n – 1
[1 + 15n + nC2 (15)2 + nC3 (15)3 + ........ nCn (15)n ] – 15n – 1

225 [ n C2  n C3 ·15  ......  n Cn · (15) n  2 ]


integer

 it is always a number divisible by 225. ]

(13) 25n + 5 – 31n – 32 is divisible by 961.
[Sol. (32)n + 1 – 31n – 32
(1 + 31)n + 1 – 31n – 32
[1 + 31(n + 1) + n + 1C2(31)2 + ...... + n + 1Cn + 1(31)n + 1 – 31n – 32]
961 [ n 1 C 2  n 1C3 · 31  ......  n 1C n 1 · (31) n 1]
  
integer

Hence divisible by 961. ]

(14) Find the coefficient of xr in
y = (x + 2)n–1 + (x + 2)n–2(x + 1) + (x + 2)n – 3(x + 1)2 + ......+ (x + 1)n–1.

  x  1   x  1 3  x  1 
n 1 
y = (x + 2)n – 1 1      .....    ;
  x  2   x  2   x  2  

Bansal Classes Page # 14

   x  1 n 
1    

  x  2   [(x  2)n  ( x  1)n ]
y = (x + 2)n – 1 = (x + 2)n = (x + 2)n – (x + 1)n
n
  x  1  ( x  2)
1  
  x  2  
 

 coefficient of xr in (x + 2)n – (x + 1)n

(2 + x)n – (1 + x)n
coefficient of xr = nCr · 2n – r – nCr
= nCr(2n – r – 1) ]

n  1 r  3 r  7 r  15  r
r n

(15) Evaluate the sum:  (  1) · C r              ......  m terms 
r 0  2   4   8   16  

n r n n r r
r n 1 r n  3 r n 7
[Sol.  ( 1) · C r     ( 1) · C r     ( 1) · C r    .........m terms
r 0  2  r 0  4  r 0 8

n n n n n n
 1  3  7 1 1 1
= 1    1    1    ....m terms =          ...m terms
 2  4  8  2  4 8

n
1 1
=    
2n
1
 
3n
1
 ....   
mn
;    ·
n

 1  1  1 2 mn 2m n  1
= mn n ]
n
2 2 2 2  2  1  1  2 (2  1)
 
2
(16) Coefficient of x2y3z4 in the expansion of (ax – by + cz)9.
[Sol. Tr + 1 = 9Cr(ax – by)9 – r (cz)r
for coefficient of z4 r=4
T5 = 9C4 (ax – by)5 c4 z4
= 9C4 {5C3 (ax)2 (– by)3 } c4 z4
= 9C4 5C3 (ax)2 (– by)3 c4 z4
= – 9C4 5C3 a2 b3 c4 x2 y3 z4
 9! 
 coefficient of x2b3c4 is   a 2 b3c 4  ]
 4!· 3!· 2! 
(17) Coefficient of x4 in the expansion of (2 – x + 3x2)6
[Sol. (2 – x + 3x2)6 = [2 + x(3x – 1)]6
Tr + 1 = 6Cr 26 – r xr(3x – 1)r
r = 2, r = 3, r = 4
r = 2,
T3 = 6C2 24 x2(3x – 1)2
here coefficient of x4 is
6C 24 × 9 = 15 × 16 × 9 = 2160
2
r = 3,
T3 = 6C3 23 x3(3x – 1)3
here coefficient of x4 is
6C 23 × x3 3(3x)(–1)2 = 6C · 8 · 9 = 20 × 72 = 1440
3 3

Bansal Classes Page # 15

 r = 4,
T3 = 6C4 22 x4(3x – 1)4
here coefficient of x4 is
6C · 22 · 1 (–1)4 = 15 × 4 × 1 = 60
4
 coefficient of x4 in (2 – x + 3x2)6 is 2160 + 1440 + 60 = 3660 ]
Important concept:

(1) If n  N then prove that the integral part of the number N = 3  7  n is an odd integer..
I

[Sol. N = 3  7 n I  integral part & F  positive fractional part. 0< F<1
F
Consider 3  7 is a positive positive proper fraction

 3  7 n will also be a positive proper fraction say F'.

consider (I + F) + (F')

I + F + F' = 3  7 n + 3  7 n
2  C0 3  C 2 3  7   ........
n n n n 2 2 
 
 I + F + F' = even integer
but I is an integer
 F + F' must also be an integer and 0 < F + F' < 2
 F + F' = 1
 I = Even integer – 1  I = odd integer ]

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

 
[Sol. Consider 2  3 which is a positive proper fraction F'

(2 + 3 )6 + (2 – 3 )6 = 2[ 6C0 26 + 6C2 24 ( 3 )2 + 6C4 22 ( 3 )4 + 6C6 ( 3 )6 ]

= 2[64 + 15 · 16 · 3 + 15 · 4 · 9 + 27]
I + F + F' = 2[1351] = 2702
again I  integer
 F + F'  integer;  F + F' = 1
 I = 2701 ]

I
(3) Let N = (2 + 3 )6 then the value of N(1 – F).
F
[Sol. F + F' = 1 where F' = (2 – 3 )6
(2 + 3 )6 (F') = [(2 + 3 ) (2 – 3 )]6 = 1 ]

Bansal Classes Page # 16

(4) Show that the integral part of the number N = ( 3 3 + 5)2n + 1 is even (n  N).
I
[Sol. N = ( 3 3 + 5)2n + 1 0<F<1
F
consider ( 3 3 – 5)  F' 0 < F' < 1
( 3 3 + 5)2n + 1 – ( 3 3 – 5)2n + 1
= 2[ 2n + 1C0 ( 3 3 )2n · 51 + 2n + 1C3( 3 3 )2n – 2 · 53.........]
I + F – F' = even integer
–1 < F – F' < 1
but again I is an integer, therefore (F – F')  I but – 1 < F < F' < 1
hence F – F' = 0
I = even integer  IF = 3 3  5  
2 n 1 2 n 1
 · 3 3 5 = 22n + 1. ]

(5) 
If N = 7  4 3 n = p +  (n  N) where p is the integral part of N and p is the
positive proper fraction then find the value of (1 – )(p + ).
[Sol. 7  4 3 n  ' positive proper fraction
and  + ' = 1
 1 –  = '
'(p + ) (7 + 4 3 )n (7 – 4 3 )n = (49 – 48) = 1 Ans. ]
Home work: Exercise-I(A) Q.1 to Q.15, Exercise-1(B) complete.
2n 2n
r
(6) If  a r ( x  2) =  br (x  3) r and ak = 1  r  n then prove that bn = 2n + 1Cn + 1.
r 0 r 0
[Sol. Let x–3=y and x–2=1+y
2n 2n
 a r (1  y)r =  br yr
r 0 r 0
In RHS bn is the coefficient of yn. and
LHS is a0 + a1(1 + y) + a2(1 + y)2 + an(1 + y)n + an + 1(1 + y)n + 1 + ...... + a2n(1 + y)2n
and an = an + 1 = an + 2 = ........ = a2n = 1
 bn is coefficient of yn in (1 + y)n + (1 + y)n + 1 + ..... + (1 + y)2n
 bn = nCn + n + 1Cn + n + 2Cn + .............. + n + nCn
bn = n + 1Cn + 1 + n + 1Cn + n + 2Cn + ...... + n + nC
n
bn = n + 2C + n + 2C + n + 3C + ...... + n + nC
n+1 n n n
bn = n + 3Cn + 1 + n + 3C
n + ................... + n + nC
n
bn = n + nCn + 1 + n + nCn
bn = 2n + 1Cn + 1

Bansal Classes Page # 17

Asking:
(i) Sum of the last ten coefficient in the expansion of (1 + x)19
[Sol. (1 + x)19 = 19C0 + 19C1x + 19C2x2 + ........ + 19C19x19
19
and 19C + 19C1 + 19C2 + ..... + 19C9 + C10  19C11  ......  19C19 = 219
0 
x

 2[19 C10  19C11  19C12  ......  19C19 ] = 219


x
x = 218 ]

(ii) Prove that nC0 + n + 1C1 + n + 2C2 + ........ + n + rCr = n + r + 1Cr.

[Sol. n + 1C0 + n + 1C1 + n + 2C2 + ........ + n + rCr
n + 2C + n + 2C2 + n + 3C3 + ........ + n + rCr
1
n + 3C n + 3C + ........ + n + rC
2+ 3 r
= n + 4C n + r n + r + 1
r–1 + Cr = Cr Hence proved. ]

1 10
(iii) S= C0 – 10C1 + 10C2 · 2 – 10C3 · 22 + ...... + 10C10 29 = ??
2
1 10
[Sol. S = [ C0 – 10C1 · 2 + 10C2 · 22 – 10C3 · 23 + ...... + 10C10 210]
2
1 1
S= (1 – 2)10  S= ]
2 2

(7) Find the sum of all the coefficient of all the integral power of 'x' in the expansion of

1  2 x 40 .
[Sol. 1  2 x 40 = 40C0 + 40C1 ( 2 x )1 + 40C2 ( 2 x )2 + ....... + 40C40 ( 2 x )40
1  2 x 40 = 40C0 + 40C1 (– 2 x )1 + 40C2 (– 2 x )2 + ...... + 40C40 (– 2 x )40
 1  2 x 40 + 1  2 x 40 = 2[40C0 + 40C2 ( 2 x )2 + ....... + 40C40 ( 2 x )40]
put x=1

340  1 40
= C0 + 40C2 · 22 + 40C4 · 24 + ...... + 40C40 · 240 ]
2
Binomial theorem for any index (negative or fractional) i.e. n  Q
|x|<1
n (n  1) 2 n (n  1)(n  2) 3
(i) (1 + x)n = 1 + nx + x  x  ........
2! 3!

n (n  1) 2 n (n  1)(n  2) 3
(ii) (1 + x)–n = 1 – nx + x  x  ........
2! 3!

Bansal Classes Page # 18

 n (n  1) 2 n (n  1)(n  2) 3
(iii) (1 – x)n = 1 – nx + x  x  ........
2! 3!

n (n  1) 2 n (n  1)(n  2) 3
(iv) (1 – x)–n = 1 + nx + x  x  ........
2! 3!

|x|<1
(1) (1 + x)–1 = 1 – x + x2 – x3 + ........... 
(2) (1 – x)–1 = 1 + x + x2 + x3 + ........... 
(3) (1 + x)–2 = 1 – 2x + 3x2 – 4x3 + ........... 
(4) (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + ................. 
(5) (1 + x)–3 = 1 – 3x + 6x2 – 10x3 ................... 
(2) (1 – x)–3 = 1 + 3x + 6x2 + 10x3 ................... 

EXAMPLES:
(1) Find the range of x for which this expansion is valid (8 + x)4/3.
4 4 4 4
 x 3  x 3 x
[Sol. (8  x ) 3  83 1 
  ; 161   ; Now < 1; | x | < 8
 8  8 8
–8<x<8 ]

(2) Find General term of (1 – x)–3. [Ans. (r + 2Cr) xr ]

3(4) x 2 3· 4 · 5 x 3
[Sol. (1 – x)–3 = 1 + 3x +   ........ 
2! 3!

3· 4 ·5........(r  1)(r  2) r 1· 2 ·3 · 4........(r  2) x r

Tr + 1 = x =
r! 2· r!

( r  2)! x r (r  2)( r  1) x r r + 2
= = = Cr xr Ans.]
2 r! 2

(3) Write the 8th term in the expansion of (1 + 2x)–1/2

 1  1  2  1  1  1  3
     1(2 x )      1   2 (2 x )
8 2  2  2  2  2
[Sol. (1 + 2x)–1/2 = 1  x      ......
2 2! 3!

 1  1  1   1  r
     1   2 ......   (r  1) (2 x )
Tr + 1 =  2  2  2   2 
r!

 1  1  1  1 
(1) r    1  2 ......  r  1 2r x r
=  2  2  2  2 
r!

Bansal Classes Page # 19

 (1) r 1·3·5.......(2r  1)  2r x r (1) r 1· 3· 5.......(2r  1)  x r
= =
2r r ! r!

(1)7 (1·3· 5.......13) x 7

 T8 = ]
7!

(1  1  x 2 ) x 1·3 x 3 1·3 ·5 · 7 x 5
(4) If 0 < x < 1 then prove that =    .......
2 2 2 · 4 6 2 · 4 · 6 · 8 10

1 1 1 x  1 x
[Sol. LHS =
2
2  2 1  x2 =
2
 1 x  1 x 2 =
2

x 1·3 x 3 1·3 ·5 · 7 x 5
=    .......
2 2 · 4 6 2 · 4 · 6 ·8 10
Approximation:
If x is so small so that its square and higher power can be neglected.
5
 2x  12
1    (4  2x )
Find the value of  3 
( 4  x )3 2

 10 x   x 1 2   3 2  x  3 2  10x  x   3x 
[Sol. 1  3   21  2   4 1   = 1   21   4 3 2 1  
      4  3  4   8 

 10 x x  3x   17 x  3x  9 x 17 x 51x 2
1   2 1   3   1   3  
 3 2  8   6  8  8 6 24
= =
8 8 8

(27 x  68x )
3 3 95x 3 95x
= 24 = 8  24 · 8 =  ]
8 8 192
Important :
Coefficient of xr in the expansion of (1 – x)–n, n  N is n + r – 1Cr . (used in P&C)
n (n  1) 2 n (n  1)(n  2) 3
Proof: (1 – x)–n = 1 + nx + x  x  ......
2! 3!

n (n  1)(n  2).......(n  r  1) x r 1· 2 · 3......(n  1)(n  1).......(n  r  1)!x r

Tr + 1 = =
r! (n  1)!r !
Tr + 1 = n + r – 1 Cr · xr
 coefficient of xr is n + r – 1C
r. ]

Bansal Classes Page # 20

EXAMPLES:
(1) Find the coefficient of x10 in (1 – x)–8.
[Hint: 8 + 10 – 1C10 = 17C10 ]

(2) Find the coefficient of x100 in (3 – 5x) (1 – x)–2

[Sol. 3 × coefficient of x100 in (1 – x)–2 – 5 × coefficient of x99 in (1 – x)–2
= 3 × 101C100 – 5 × 100C99 = 3 · 101C1 – 5 · 100C1
= 3 × 101 – 5 × (100)
= 303 – 500 = – 197 ]

BINOMIAL SERIES
1 1 1· 3 1 1· 3· 5 1
(3) Find the sum: S = 1 – ·  ·  ·  .........
2 2 2 · 4 2 2 2 · 4 · 6 23

n (n  1) 2
[Sol. (1 + x)n = 1 + nx + x  .......
2!

1 n (n  1) 2 3
nx = – ....(1) and x = ....(2)
4 2! 32

n 1 3
(2)  (1)2  =
2n 2
 n – 1 = 3n  n = – 1/2,  x = 1/2
1 2 1 2
 1 3 2
 S = 1      Ans. ]
 2 2 3

1 1· 3 1· 3· 5
(4) If x =    ........ find the value of x2 – 2x = ?
3 3 · 6 3· 6 · 9
or find the value of x2 + 2x = ?.
1 1· 3 1· 3· 5
[Sol. y = x + 1 = 1 +    ........
3 3 · 6 3· 6 · 9

n (n  1) 2
 let y = (1 + z)n = 1 + nz + z  .......
2!

1 n (n  1) 2 1· 3
nz = ....(1) and z = ....(2)
3 2! 18

(n  1) 3 9
(2)  (1)2  = ×  n – 1 = 3n;  n = – 1/2
2n 18 2
and n = 1/2,  z = – 2/3
1 2 1 2
 2 1
 y = 1      3
 3  3

Bansal Classes Page # 21

  x+1= 3;  (x + 1)2 = 3
x2 + 2x – 2 = 0 hence proved. ]

EXPONENTIAL SERIES:

x x 2 x3
ex =1+    .........
1! 2! 3!

x x 2 x3 x 4
e–x = 1    .........
1! 2! 3! 4!

 x2 x4 
ex + e–x 2
= 1   ......... 
2 ! 4! 
 

 x x3 x5 
and ex – e–x = 2    ......... 

 1! 3! 5! 
put x=1
1 1 1
e=1+    .........
1! 2! 3!
EXAMPLES:
2 4 6 8
(1) Find S = 1!  3!  5!  7!  .....

[Sol. We have
1 1 1 1 1 1 1
S = 1 + 1!  2!  3!  4!  5!  6!  7 !  .....  

1 1 1 1
         ....
 2 ! 3 !  4 ! 5 !

4 6
expansion of ax = ex ln a 2 + 3 !  5 ! + .............

( x ln a ) ( x ln a ) 2 ( x ln a )3
 ax = ex ln a =1+ 1! + + + ...... ]
2! 3!

(1  22 ) 2 (1  23 ) 3
(2) Find (1 + 2) ln 2 + ln 2 + ln 2 + ........ 
2! 3!

 2 ln 2 (2 ln 2)2 (2 ln 2)3   ln 2 ln 2 2 ln 3 2 

[Sol.  1!    .......  +     ....... 
2! 3!   
   1! 2! 3! 
e2 ln 2 – 1 + eln 2 – 1
4 – 1 + 2 – 1 = 6 – 2 = 4 Ans. ]

Bansal Classes Page # 22

(3) If  and  are roots of the quadratic equation 3x2 – 6x + 2 = 0 then find

   2 3    2 3 
1     
 ...... 1     ...... 
 2 2!·4 3!·8   2 2!·4 3!·8 
  

 ( 2) ( 2)2 ( 2)3   ( 2) ( 2) 2 ( 2)3 


[Sol. 1     ......  1     ...... 
1! 2! 3!  1! 2! 3! 
  
  2
e/2 · e/2 = e 2 = e2 = e Ans. ]

1  2 1  2  2 2 1  2  2 2  23
(4)(a) Find 1 +   + ...... 
2! 3! 4!

n3
(b)  n!
n 1

[Sol.

1  2  2 2  .....2 n 1 2n  1 2n  1
(a) Tn = = (2  1) n ! 
n! n!

2n  
1  2 22 23   1 1 1 
S= 
n!
–  n! = 1  1!  2!  3!  .....  – 1  1!  2!  3!  ..... 
n1 n1    
= e2 – e Ans.

n3 n2 (n 2  1)  1 (n  1) 1 (n  2)  3 1
(b)  n! = (n  1)! (n  1)!  (n  2)!  (n  1)! = (n  2)!  (n  1)!

n 1

1 3 1
= (n  3)!  (n  2)!  (n  1)! = 5e]

LOGARITHMIC SERIES : (Not in syllabus)

Result

 x x 2 x3 x 4 
(1) ln(1 + x) =      .......  ;
 –1 x1
1 2 3 4 
[Proof: y = 1 – x + x2 – x3 + x4 – ........ (diff. coefficient of)
1
y=
1 x
x x
1
 y dx   1  x dx
0 0

x
2
 (1  x  x  x 3  x 4  ......) dx = ln (1 + x)
0

Bansal Classes Page # 23

  x x 2 x3 x 4 
and ln(1 – x) = –      .......  ; – 1  x < 1
1 2 3 4 
 
1 1 1 1
Note: (a) 1      ........  ln 2
2 3 4 5

 x x3 x5 
ln(1 + x) – ln(1 – x) = 2     ....... 

1 2 3 

1 x   x x3 x5 
ln   = 2    .......  ]
1 x  1 2 3 
 
EXAMPLES:
1 1 1
(1) Find the value of S =    ...... .
2·3 4·5 6· 7

1 1 1 1 1 1
[Sol. S =       ......
2 3 4 5 6 7

 1 1 1 1 1 1 
S = – 1        ...... 
 2 3 4 5 6 7 
S = – (ln 2 – 1);  S = 1 – ln 2 ]

1 1 1 1
(2) Find the value of S = 2     ........ .
2 · 2 2 3 · 23 4 · 2 4

1 (1 2) 2 (1 2)3 (1 2) 4
[Sol.     ........
2 2 3 4

  3
ln 1   = ln = ln 3 – ln 2 ]
 2 2

Bansal Classes Page # 24