NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 8
Chapter Name : Binomial Theorem

Exercise 8.1

Q1 Expand the expression (1 − 2x) 5

5
(1 − 2x)
5 5 5 4 5 3 2 5 2 3 5 1 4 5 5
= C0 (1) − C1 (1) (2x) + C2 (1) (2x) − C3 (1) (2x) + C4 (1) (2x) − C3 (2x)
2 3 4 2
= 1 − 5(2x) + 10 (4x ) − 10 (8x ) + 5 (16x ) − (32x )
2 3 4 5
= 1 − 10x + 40x − 80x + 80x − 32x

Page : 166 , Block Name : Exercise 8.1

5

Q2 Expand the expression (
2 x
− )
x 2

5 3 4 3 2
2 x 5 2 5 2 x 5 2 x
( − ) = C0 ( ) − C1 ( ) ( ) + C2 ( ) ( )
x 2 x x 2 x 2

2 3 4 5
3 2 x 3 2 x 3 x
− C3 ( ) ( ) + C4 ( )( ) − C5 ( )
x 2 x 2 2

2 4 5
32 16 x 8 x 4 2 x x
= 5
− 5( )( ) + 10 ( 3
)( ) − 10 ( 2
) + 5( )( ) −
x x4 2 x 4 x x 16 32

3
32 40 20 5 3 x
= 5
− 3
+ − 5x + x −
x x x 8 32

Page : 166 , Block Name : Exercise 8.1

Q3 Expand the expression (2x − 3) 6

Answer. By using Binomial theorem, the expression (2x − 3) can be expressed as 6

6 6 6 6 5 6 4 2 4 3 3
(2x − 3) = C0 (2x) − C1 (2x) (3) + C1 (2x) (3) − C3 (2x) (3)
6 5 4 3
= 64x − 6 (32x ) (3) + 15 (16x ) (9) − 20 (8x ) (27)

2
+ 15 (4x ) (81) − 6(2x)(243) + 729

6 5 4 3 2
= 64x − 576x + 2160x − 4320x + 4860x − 2916x + 729

Page : 166 , Block Name : Exercise 8.1

5

Q4 Expand the expression (
x 1
+ )
3 x

5

Answer. By using Binomial Theorem, the expression ( can be expand as
x 1
+ )
3 x

5 5 4 3 2
x 1 5 x 3 x 1 3 x 1
( + ) = C0 ( ) + C1 ( ) ( ) + C2 ( ) ( )
3 x 3 3 x 3 x

5 4 3 2
x x 1 x 1 x 1 x 1 1
= + 5( )( ) + 10 ( )( 2
) + 10 ( )( 3
) + 5( )( 4
) + 5
243 81 x 27 x 9 x 3 x x

5 3
x 5x 10x 10 5 1
= + + + + 3
+ 3
243 81 27 9x 3x x

Page : 167 , Block Name : Exercise 8.1

6

Q5 Expand the expression (x + 1

x
)

6 2
1 6 6 6 ′ 1 6 4 1
(x + ) = C0 (x) + C1 (x) ( ) + C2 (x) ( )
x x x

3 4 5 6
6 3 1 6 2 1 6 1 6 1
+ C3 (x) ( ) + C4 (x) ( ) + C3 (x)( ) + C6 ( )
x x x x

4 3 1 4 1 3 1 2 1 1 1
= x + 6(x) ( ) + 15(x) ( ) + 20(x) ( ) + 15(x) ( ) + 6(x) ( ) +
x 2 3 4 5 6
x x x x x

6 4 2 15 6 1
= x + 6x + 15x + 20 + + +
2 4 6
x x x

Page : 167 , Block Name : Exercise 8.1

Q6 Using binomial theorem, evaluate (96) 3

96 can be expressed as the sum or difference of two numbers whose powers are easier

to calculate and then, binomial theorem can be applied.

It can be written that, 96 = 100 − 4
3 3
∴ (96) = (100 − 4)
3 3 3 2 3 2 3 3
= Cα (100) − C1 (100) (4) + Cz (100)(4) − C3 (4)

3 2 2 3
= (100) − 3(100) (4) + 3(100)(4) − (4)

= 1000000 − 120000 + 4800 − 64

= 884736

Page : 167 , Block Name : Exercise 8.1

Q7 Using binomial theorem, evaluate (102) 5

102 can be expressed as the sum or difference of two numbers whose powers are easier

to calculate and then, Binomial Theorem can be applied.

It can be written that, 102 = 100 + 2
5 5
∴ (102) = (100 + 2)
′
∗ 5 4 5 3 2 5 2 3
= C0 (100) + C1 (100) (2) + C2 (100) (2) + C3 (100) (2)
5 4 3 2 2 3 4 5
= (100) + 5(100) (2) + 10(100) (2) + 10(100) (2) + 5(100)(2) + (2)

= 1000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

Page : 167 , Block Name : Exercise 8.1

Q8 Using binomial theorem, evaluate (101) 4

101 can be expressed as the sum or difference of two numbers whose powers are easier

to calculate and then, Binomial Theorem can be applied.

It can be written that, 101 = 100 + 1
4 4
∴ (101) = (100 + 1)

4 4 4 3 4 2 2 4 2 4 4
= C0 (100) + C1 (100) (1) + C2 (100) (1) + C3 (100)(1) + C4 (1)

4 3 2 4
= (100) + 4(100) + 6(100) + 4(100) + (1)

= 100000000 + 400000 + 60000 + 400 + 1

= 104060401

Page : 167 , Block Name : Exercise 8.1

Q9 Using binomial theorem, evaluate (99) 5

99 can be written as the sum or difference of two numbers whose powers are easier to

calculate and then, Binomial Theorem can be applied.

It can be written that, 99 = 100 − 1
5 5
. (99) =(100 − 1)

5 5 5 4 5 3 2 5 2 3
= C0 (100) − C1 (100) (1) + C2 (100) (1) − C3 (100) (1)

3 4 5 5
+ C4 (100)(1) − C5 (1)
5 4 3 2
= (100) − 5(100) + 10(100) − 10(100) + 5(100) − 1

= 10000000000 − 500000000 + 10000000 − 100000 + 500 − 1

= 10010000500 − 500100001

= 9509900499

Page : 167 , Block Name : Exercise 8.1

Q10 Using Binomial Theorem, indicate which number is larger (1.1) 10000
or 1000

10000
By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1) can

be obtained as
10000 10000
(1.1) = (1 + 0.1)

10000
= (1 + 0.1) C1 (1.1) + Other positive terms

= 1 + 10000 × 1.1 + Other positive terms

= 1 + 11000 + Other positive terms

> 1000
10000
(1.1) > 1000

Page : 167 , Block Name : Exercise 8.1

Q11 Find (a + b) 4 4
− (a − b) . Hence, evaluate (√3 + √2)
4
− (√3 − √2)
4

Answer. Useing Bionomial Theorm, the wxpression (a + b) 4
and (a − b) ,
4
can be expanded as
4 4 4 + 3 + 2 2 4 3 4 4
(a + b) = C0 a + C1 a b + C2 a b + C3 ab + C4 b

4 4 4 4 3 4 2 2 4 3 4 4
(a − b) = C0 a − C1 a b + C2 a b − C3 ab + C4 b

4 4 4 4 4 3 4 2 2 4 3 4 4
∴ (a + b) − (a − b) = C0 a + C1 a b + C2 a b + C3 ab + C4 b

4 4 4 3 4 2 2 4 3 ⋅ 4
− [ C0 a − C1 a b + C2 a b − C3 ab + C4 b ]

4 3 4 3 3 3
=2 ( C1 a b + C3 ab ) = 2 (4a b + 4ab )

2 2
=8ab (a + b )

By putting a = √3 and b = √2, we obtain
4 4 2 2
(√3 + √2) − (√3 − √2) = 8(√3)(√2) {(√3) + (√2) }

= 8(√6){3 + 2} = 40√6

Page : 167 , Block Name : Exercise 8.1

Q12 Find (x + 1) 6
+ (x − 1)
6
⋅ Hence or otherwise evaluate (√2 + 1)
6
+ (√2 − 1)
6

6 6
Using Binomial Theorem, the expressions, (x + 1) and (x − 1) , can be expanded as

6 6 6 6 3 6 4 6 3 6 2 6 6
(x + 1) = C0 x + C1 x + C2 x + C3 x + C4 x + C5 x + C6

6 6 6 6 5 6 4 6 3 6 2 6 6
(x − 1) = C0 x − C1 x + C2 x − C3 x + C4 x − C5 x + C6
6 6 6 6 6 4 6 2 6
∴ (x + 1) + (x − 1) = 2 [ C0 x + C2 x + C4 x + C6 ]

6 4 2
= 2 [x + 15x + 15x + 1]

By putting x = √2, we obtain
6 6 ∘ 4 2
(√2 + 1) + (√2 − 1) = 2 [(√2) + 15(√2) + 15(√2) + 1]

= 2(8 + 15 × 4 + 15 × 2 + 1)

= 2(8 + 60 + 30 + 1)

= 2(99) = 198

Page : 167 , Block Name : Exercise 8.1

Q13 Show that 9 n+1
− 8n − 9 is divisible by 64, whenever n

n+1
In order to show that 9 − 8n − 9 is divisible by 64, it has to be proved that,

n+1
9 − 8n − 9 = 64k, where k is some natural number

By Binomial Theorem,

m m m m 2 m m
(1 + a) = C0 + C1 a + C2 a + … + Cm a

For a = 8 and m = n + 1, we obtain

n+1 n+1 n+1 n+1 2 n+1 n+1
(1 + 8) = C0 + C1 (8) + C2 (8) + … + Ca+1 (8)
n+1 2 n+1 n+1 n+1 n−1
⇒ 9 = 1 + (n + 1)(8) + 8 [ C2 + C3 × 8 + … + Cn+1 (8) ]

n+1 n+1 11 n+1 n−1
⇒ 9 = 9 + 8n + 64 [ C2 + C3 × 8 + … + C0+1 (8) ]

n+1 n+1 n+1 n+1 n−1
⇒ 9 − 8n − 9 = 64k, where k = C2 + C3 × 8 + … + Cn+1 (8) is a natural number

n+1
Thus, 9 − 8n − 9 is divisible by 64, whenever n is a positive integer.

Page : 167 , Block Name : Exercise 8.1

Q14 Prove that ∑
n rn n
3 C, = 4
r=0

Answer. By Binomial Theorem,
n n−r r n
∑ Ct a b = (a + b)
i=0

By putting b = 3 and a = 1 in the above equation, we obtain
n n−1 r n
∑ Ct (1) (3) = (1 + 3)
r=0

n n n
⇒ ∑ 3 Cr = 4
t=0

Hence, proved.

Page : 167 , Block Name : Exercise 8.1

Exercise 8.2

Q1 Find the coef cient of x ln(x + 3) 5 8

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−t t
Tr+1 = Cr a b
5 th 8
Assuming that x occurs in the (r + 1) term of the expansion (x + 3) , we obtain

8 3−t r
Tr+1 = Cr (x) (3)
5
Comparing the indices of x in x and in Tr+1, we obtain

r = 3

Thus, the coef cient of x 5 3 3 8! 3 8⋅7⋅6⋅5! 3
is C3 (3) = × 3 = ⋅ 3 = 1512
3!5! 3⋅2.5!

Page : 171 , Block Name : Exercise 8.2

Q2 Find the coef cient of a b ln(a − 2b) 5 7 12

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−1 r
Tr+1 = Cr a b
5 7 th 12
a b occurs in the (r + 1) term of the expansion (a − 2b) ,
12 12−r ′ 12 ′ 12−T r
Tr+1 = Cr (a) (−2b) = Cr (−2) (a) (b)

Thus the coef cient of a b is 5 7

12 7 12! 7 12⋅11⋅10⋅9⋅8.7! 7
C7 (−2) = − ⋅ 2 = − ⋅ 2 = −(792)(128) = −101376
715! 5⋅4⋅3⋅2.7!

Page : 171 , Block Name : Exercise 8.2

6
Q3 Write the general term in the expansion of (x 2
− y)

th
It is known that the general term Tr+1 {which is the (r + 1) term } in the binomial

n n n−τ r
expansion of (a + b) is given by Tr+1 = Cr a b

2 6
Thus, the general term in the expansion of (x − y ) is

6−r
6 2 t r 12−2r r
Tr+1 = C1 (x ) (−y) = (−1) Cr ⋅ x y

Page : 171 , Block Name : Exercise 8.2

12
Q4 Write the general term in the expansion of (x 2
− yx) ,x ≠ 0

th
It is known that the general term Tr+1 { which is the (r + 1) term } in the binomial

n n n−τ r
expansion of (a + b) is given by Tr+1 = Cr a b
12
Thus the general term of expansion of(x 2
− yx) is
12−t
12 2 r 12 2+2r r r 24−r r
Tr+1 = Cr (x ) (−yx) = (−1)r Cr ⋅ x ⋅ y ⋅ xr = (−1) Cr ⋅ x ⋅ y

Page : 171 , Block Name : Exercise 8.2

Q5 Find the 4 th
term in the expansion of (x − 2y)
12

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−1 ′
Tr+1 = C1 a b
th 12
Thus, the 4 term in the expansion of (x − 2y) is

12 12−3 3 3 12! 0 3 4 12⋅11⋅10 3 4 3 9 3
T4 = T7+1 = C3 (x) (−2y) = (−1) ⋅ ⋅ x ⋅ (2) ⋅ y = − ⋅ (2) x y = −1760x y
3!9! 3⋅2

Page : 171 , Block Name : Exercise 8.2

18

Q6 Find the 13
th 1
term in the expansion of (9x − ) ,x ≠ 0
3√x

th n
(r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b)
12

1
18 18−12
T13 = T12+1 = C12 (9x) (− )
3√x

12
12
18! 1 1
12 6 6
= (−1) (9) (x) ( ) ( )
12!6! 3 √x

18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12! 1 1
6 12
= ⋅ x ⋅ ( ) ⋅ 3 ( )
6 12
12!6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 x 3

= 18564

Page : 171 , Block Name : Exercise 8.2

7

Q7 Find the middle terms in the expansions of (3 −
3
x
)
6

Answer. It is known that in the expansion of (a + b) , if n if n is odd then there are two middle terms, namely
n

th t

Term
n+1 n+1
( ) term and ( + 1)
2 2

3
3 9
x 7! x
7 1−3 3 4
T4 = T3+1 = C3 (3) (− ) = (−1) ⋅ 3 ⋅
3
6 3!4! 6

7 ⋅ 6 ⋅ 5.4! 1 105
4 9 9
= − ⋅ 3 ⋅ ⋅ x = − x
3 3
3 ⋅ 2.4! 2 ⋅ 3 8
4
3 12
7 7−4 x 4 7! 3 x
T5 = T4,1 = C4 (3) (− ) = (−1) (3) ⋅
6 4
4!3! 6
3
7⋅6⋅5⋅4! 3 12 35 12
= ⋅ ⋅ x = x
4 4 48
4!.3⋅2 2 ⋅3
7
3
x 105 9 35 12
(3 − ) are − x and x
6 8 48

Page : 171 , Block Name : Exercise 8.2

10

Q8 Find the middle terms in the expansion of (
x
+ 9y)
3

th
n n
It is known that in the expansion (a + b) , if n is even, then the middle term is ( + 1)
2

term.
10−5
5
10 x 5 10! x 5 5
T6 = T5+1 = C5 ( ) (9y) = ⋅ 5
⋅ 9 ⋅ y
3 5!5! 3

10.9⋅8⋅7⋅6.5! 1 3 5 5
= ⋅ ⋅ ⋅x y
5
5⋅4⋅3⋅2.5! 3
5 5 5 5 5
= 252 × 3 ⋅ x ⋅ y = 61236x y

Page : 171 , Block Name : Exercise 8.2

Q9 In the expansion of (1 + a) m+n
, prove that coefficients of a
m
and a
n

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−τ r
Tr+1 = Cr a b
m th m+n
Assuming that a occurs in the (r + 1) term of the expansion (1 + a) , we obtain

m+n m+n−r r m+n r
Tr+1 = Cr (1) (a) = Cr a
m
Therefore, the coefficient of a is

(m+n)! (m+n)!
m + nCm = =
m!(m+n−m)! m!n!

n th m+n
Assuming that a occurs in the (k + 1) term of the expansion (1 + a) , we obtain

m+n m+n−k k m+n k
Tk+1 = Ck (1) (a) = Ck (a)
n
Therefore, the coefficient of a is

(m+n)! (m+n)!
m + nCn = =
n!(m+n−n)! n!m!

m n
Thus, from (1) and (2), it can be observed that the coefficients of a and a in the expansion of

m+n
(1 + a) are equal.

Page : 171 , Block Name : Exercise 8.2

Q10
th th th n
The coefficients of the (r − 1) ,r and (r + 1) terms in the expansion of (x + 1)

are in the ratio 1 : 3 : 5. Find n and r.

th n
It is known that (k + 1) term, (Tk+1 ) , in the binomial expansion of (a + b) is given by

n n−k k
Tk+1 = Ck a b
th n n n−(r−2) (r−2) n n−r+2
(r − 1) term in the expansion of (x + 1) is Tr−1 = Cr−2 (x) (1) = Cr−2 x
th n n n−(r−1) (r−1) n n−r+1
r term in the expansion of (x + 1) is Tr = Cr−1 (x) (1) = Cr−1 x
th th th n
Therefore, the coefficients of the (r − 1) ,r , and (r + 1) terms in the expansion of (x + 1)

n n n
are Cr−2 , Cr−1 , and Cr respectively.

since these coef cients are in the ratio } 1 : 3 : 5, we obtain
n n
Cr−2 1 Cr−1 3
n
= and n
=
Cr−1 3 Cr 5

n
Cr−2 n! (r − 1)!(n − r + 1)!
= ×
n
Cr−1 (r − 2)!(n − r + 2)! n!

(r − 1)(r − 2)!(n − r + 1)!
=
n!

(r − 1)(r − 2)!(n − r + 1)!
=
(r − 2)!(n − r + 2)(n − r + 1)!

r − 1
=
n − r + 2
r−1 1
∴ =
n−r+2 3

⇒ 3r − 3 = n − r + 2

⇒ n − 4r + 5 = 0
n
Cr−1 = n! r!(n−r)! r(r−1)!(n−r)!

n
= × =
Cr (r−1)!(n−r+1) n! (r−1)!(n−r+1)(n−r)!

r
=
n−r+1

r 3
∴ =
n−r+1 5

⇒ 5r = 3n − 3r + 3

⇒ 3n − 8r + 3 = 0

Multiplying (1) by 3 and subtracting it from (2), we obtain

Putting the value of r in (1), we obtain

n − 12 + 5 = 0

⇒ n = 7

Thus, n = 7 and r = 3

Page : 171 , Block Name : Exercise 8.2

Q11
n 2
Prove that the coefficient of x in the expansion of (1 + x) istwice the coefficient

n 2n−1
of x in the expansion of (1 + x) .

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−t r
Tr+1 = Cr a b
n th 2n
Assuming that x occurs in the (r + 1) term of the expansion of (1 + x) , we obtain

2n 2n−r r 2n r
Tr+1 = Cr (1) (x) = Cr (x)
n 2n
Therefore, the coefficient of x in the expansion of (1 + x) is

(2n)! (2n)! (2n)!
2n
Cn = = =
n!(2n−n)! n!n! (n!)2

n th 2n−1
Assuming that x occurs in the (k + 1) term of the expansion (1 + x) , we obtain

2n−1 2n−1−k k 2n−1 k
Tk+1 = Ck (1) (x) = Ck (x)
n 2n−1
x in the expansion of (1 + x)

(2n − 1)! (2n − 1)!
2n − 1 = =
n!(2n − 1 − n)! n!(n − 1)!

2n ⋅ (2n − 1)! (2n)! 1 (2n)!
= = = [ ]
2
2n. n!(n − 1)! 2. n!n! 2 (n!)

From (1) and (2), it is observed that

1 2n 2n−1
( Cn ) = Cn
2

2n 2n−1
⇒ Cn = 2 ( Cn )
n 2n n
Therefore, the coefficient of x in the expansion of (1 + x) is twice the coefficient of x in the

2n−1
expansion of (1 + x) .

Hence, proved.

Page : 171 , Block Name : Exercise 8.2

Q12
2
Find a positive value of m for which the coefficient of x in the expansion

n
(1 + x) is 6.

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−t r
Tr+1 = Cr a b
2 th m
Assuming that x occurs in the (r + 1) term of the expansion (1 + x) , we obtain

m m−r r m r
Tr+1 = Cr (1) (x) = Cr (x)
2
Comparing the indices of x in x and in Tr+1 , we obtain

r = 2

2 m
Therefore, the coefficient of x is C2

2 m
It is given that the coefficient of x in the expansion (1 + x) is 6
m
∴ C2 = 6

m!
⇒ = 6
2!(m−2)!

m(m−1)(m−2)!
⇒ = 6
2×(m−2)!

⇒ m(m − 1) = 12

2
⇒ m − m − 12 = 0

2
⇒ m − 4m + 3m − 12 = 0

⇒ m(m − 4) + 3(m − 4) = 0

⇒ (m − 4)(m + 3) = 0

⇒ (m − 4) = 0 or (m + 3) = 0

⇒ m = 4 or m = −3
2
Thus, the positive value of m, for which the coefficient of x in the expansion

m
(1 + x) is 6, is 4 .

Page : 171 , Block Name : Exercise 8.2

Miscellaneous Exercise

Q1
n
Find a, b and n in the expansion of (a + b) if the first three terms of the expansion

are 729, 7290 and 30375, respectively.

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−t r
Tr+1 = Cr a b

The first three terms of the expansion are given as 729, 7290, and 30375 respectively.

Therefore, we obtain

n n−0 0 n
T1 = C0 a b = a = 729

n n−2 1 n−1
T2 = C1 a b = na b = 7290

n(n−1)
n n−2 2 n−2 2
T3 = C2 a b = a b = 30375 … (3)
2

Dividing (2) by (1), we obtain

n−1 n 7290
na ba =
729

⇒ nba = 10

Dividing (3) by (2), we obtain

n−2 2 n−1 30375
n(n − 1)a b 2na b =
7290

30375
⇒ (n − 1)b2a =
7290

30375×2 25
⇒ (n − 1)ba = =
7290 3

b 25
⇒ nba − =
a 3

25
⇒ 10 − ba =
3

25 5
⇒ ba = 10 − =
3 3

From (4) and (5), we obtain

5
n ⋅ = 10
3

⇒ n = 6

Substituting n = 6 in equation (1), we obtain

6
a = 729

6
⇒ a = √729 = 3

From (5), we obtain

b 5
= ⇒ b = 5
3 3

Thus, a = 3, b = 5, and n = 6

Page : 175 , Block Name : Miscellaneous Exercise

Q2 Find a if the coefficients of x 2
and x
3
in the expansion of (3 + ax)
9

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−T r
Tr+1 = Cr a b
2 th 9
Assuming that x occurs in the (r + 1) term in the expansion of (3 + ax) , we obtain

9 9−t r 9 9−r ′ r
Tr+1 = Cr (3) (ax) = Cr (3) a x

2
Thus, the coefficient of x is

9 9−2 2 9! 7 2 7 2
C2 (3) a = (3) a = 36(3) a
2!7!

3 th 9
Assuming that x occurs in the (k + 1) term in the expansion of (3 + ax) , we obtain

9 9−k k 9 9−k k k
Tk+1 = Ck (3) (ax) = Ck (3) a x
3
Thus, the coefficient of x is

9 9−3 3 9! 6 3 6 3
C3 (3) a = (3) a = 84(3) a
3!6!

6 3 7 2
84(3) a = 36(3) a

⇒ 84a = 36 × 3

36×3 104
⇒ a = =
84 84

9
⇒ a =
7

9
Thus, the required value of a is
7

Page : 175 , Block Name : Miscellaneous Exercise

Q3 Find the coef cient of x 5
in the product (1 + 2x) (1 − x)
6 7

6 6 6 6 2 6 3 6 4
(1 + 2x) = C0 + C1 (2x) + C2 (2x) + C3 (2x) + C4 (2x)

6 5 6 6
+ C5 (2x) + C6 (2x)

2 3 4 5 6
= 1 + 6(2x) + 15(2x) + 20(2x) + 15(2x) + 6(2x) + (2x)

2 3 4 5 6
= 1 + 12x + 60x + 160x + 240x + 192x + 64x
7 7 7 7 2 7 3 7 4
(1 − x) = C0 − C1 (x) + C2 (x) − C3 (x) + C4 (x)

7 5 7 6 7 7
− C5 (x) + C6 (x) − C7 (x)

2 3 4 5 6 7
= 1 − 7x + 21x − 35x + 35x − 21x + 7x − x
6 7
∴ (1 + 2x) (1 − x)

2 3 4 5 6 2 3 4 5 6 7
= (1 + 12x + 60x + 160x + 240x + 192x + 64x ) (1 − 7x + 21x − 35x + 35x − 21x + 7x − x )

Page : 175 , Block Name : Miscellaneous Exercise

Q4
n n
If a and b are distinct integers, prove that a − b is a factor of a − b , whenever

n is a positive integer.

n n
In order to prove that (a − b) is a factor of (a − b ) , it has to be proved that

n n
a − b = k(a − b), where k is some natural number

It can be written that, a = a − b + b
n n n
∴ a = (a − b + b) = [(a − b) + b]
n n n n−1 n n−1 n n
= C0 (a − b) + C1 (a − b) b + … + Cn−1 (a − b)b + Cn b

n n n−1 n n−1 n
= (a − b) + C1 (a − b) b + … + Cn−1 (a − b)b + b
n n n−1 n n−2 n n−1
⇒ a − b = (a − b) [(a − b) + C1 (a − b) b + … + Cn−1 b ]

′′ n
⇒ a − b = k(a − b)

n−1 n n−2 n n−1
where, k = [(a − b) + C1 (a − b) b + … + Cn−1 b ] is a natural number

n n
This shows that (a − b) is a factor of (a − b ) , where n is a positive integer.

Page : 175 , Block Name : Miscellaneous Exercise

Q5 Evaluate
6 6
(√3 + √2) − (√3 − √2)

6 6
Firstly, the expression (a + b) − (a − b) is simplified by using Binomial Theorem.

This can be done as

6 6 6 6 5 6 4 3 6 3 3 6 2 4 6 1 5 6 6
(a + b) = C0 a + C1 a b + C2 a b + C3 a b + C4 a b + C5 a b + C6 b

6 5 4 2 3 3 2 4 5 6
= a + 6a b + 15a b + 20a b + 15a b + 6ab + b
6 6 6 6 5 6 4 2 6 3 3 6 2 4 6 1 5 6 6
(a − b) = C0 a − C1 a b + C2 a b − C3 a b + C4 a b − C5 a b + C6 b

6 5 4 2 3 3 2 4 5 6
= a − 6a b + 15a b − 20a b + 15a b − 6ab + b

Putting a = √3 and b = √2, we obtain
6 6 5 3 3 5
(√3 + √2) − (√3 − √2) = 2 [6(√3) (√2) + 20(√3) (√2) + 6(√3)(√2) ]

= 2[54√6 + 120√6 + 24√6]

= 2 × 198√6

= 396√6

Page : 175 , Block Name : Miscellaneous Exercise

Q6 Find the value of
4 4
2 2 2 2
(a + √a − 1) + (a − √a − 1)

4 4
Firstly, the expression (x + y) + (x − y) is simplified by using Binomial Theorem.

This can be done as

4 4 4 4 3 4 2 2 + 3 4 4
(x + y) = C0 x + C1 x y + C2 x y + C3 xy + C4 y

4 3 2 2 3 4
= x + 4x y + 6x y + 4xy + y
4 4 4 4 3 4 2 2 4 3 4 4
(x − y) = C0 x − C1 x y + C2 x y − C3 xy + C4 y

4 3 2 2 3 4
= x − 4x y + 6x y − 4xy + y

4 4 4 2 2 4
∴ (x + y) + (x − y) = 2 (x + 6x y + y )

2 2
Putting x = a and y = √a − 1, we obtain
4 4 2 4
2 2 2 2 4 2
2 2 2 2
(a + √a − 1) + (a − √a − 1) = 2 [(a ) + 6(a ) (√a − 1) + (√a − 1) ]

2
8 4 2 2
= 2 [a + 6a (a − 1) + (a − 1) ]

8 6 4 4 2
= 2 [a + 6a − 6a + a − 2a + 1]

8 6 4 2
= 2 [a + 6a − 5a − 2a + 1]

8 6 4 2
= 2a + 12a − 10a − 4a + 2

Page : 175 , Block Name : Miscellaneous Exercise

Q7 Find an approximation of (0.99) using the rst three terms of its expansion 5

0.99 = 1 − 0.01
5 5
∴ (0.99) = (1 − 0.01)

5 5 5 4 5 3 2
= C0 (1) − C1 (1) (0.01) + C2 (1) (0.01)

2
= 1 − 5(0.01) + 10(0.01)

= 1 − 0.05 + 0.001

= 0.951

= 0.951

Page : 175 , Block Name : Miscellaneous Exercise

Q8
Find n, if the ratio of the fifth term from the beginning to the fifth term from the
n
4 1
end in the expansion of (√2 + ) is √6 : 1
4
√3

th
n n
It is known that in the expansion (a + b) , if n is even, then the middle term is ( + 1)
2

term.
4 4
n 4
n
(√2) (√2) n!
n 4 n−1 1 n 1 n 1 4 n
C4 (√2) ( ) = C4 ⋅ = C4 ⋅ = (√2)
4 3 3
4
√ 3 (√2)
4 2 6.4!(n−4)!

n−4 4 4
1 (√3) 3 6n! 1
n 4 4 n n
Cn−4 (√2) ( ) = Cn−1 ⋅ 2 ⋅ = Cn−1 ⋅ 2 ⋅ = ⋅
4 4
n 4 n 4 n
√3 (√3) (√3) (n−4)!4! (√3)

n! 4 n 6n! 1
(√2) : ⋅ = √6 : 1
4
6.4!(n−4)! (n−4)!!4! (√3)n

4 n
(√2) 6
⇒ : = √6 : 1
6 4
n
(√3)

4 n 4 n
(√2) (√3)
⇒ × = √6
6 6

4 n
⇒ (√6) = 36√6
n 5

⇒ 64 = 62

n 5
⇒ =
4 2

5
⇒ n = 4 × = 10
2

Thus, the value of n is 10.

Page : 175 , Block Name : Miscellaneous Exercise

4

Q9 Expand using binomial theorem (1 +
x 2
− ) ,x ≠ 0
2 x

th n
It is known that (r + 1) term, (Tr+1 ) , in the binomial expansion of (a + b) is given by

n n−1 ′
Tr+1 = Ci a b
m th m+n
Assuming that a occurs in the (r + 1) term of the expansion (1 + a) , we obtain

m+a m+a−1 r m+n r
Tr+1 = Cr (1) (a) = Cr a
m
Therefore, the coefficient of a is

(m+n)! (m+n)!
m + am = =
m!(m+n−m)! m!n!

n
Therefore, the coefficient of a is

(m+n)! (m+n)!
m + nCn = =
n!(m+n−n)! n!m!

(m+n)! (m+n)!
m+n
Cn = =
n!(m+n−n)! n!m!

m n
Thus, from (1) and (2), it can be observed that the coefficients of a and a in the

m+n
expansion of (1 + a) are equal.

Page : 176 , Block Name : Miscellaneous Exercise

3
Q10 Find the expression (3x 2
− 2ax + 3a )
2
using binomial theorem

th n
It is known that (k + 1) term, (Tk+1 ) , in the binomial expansion of (a + b) is given b

n n−k k
Tk+1 = C1 a b
th n
Therefore, (r − 1) term in the expansion of (x + 1) is

n n−(r−2) (r−2) n n−r+2
Tr−1 = C1−2 (x) (1) = Cr−2 x
3
2 2
[(3x − 2ax) + 3a ]

3 2 2 3
3 2 3 2 2 3 2 2 3 2
= C0 (3x − 2ax) + C1 (3x − 2ax) (3a ) + C2 (3x − 2ax) (3a ) + C3 (3a )

3
2 4 3 2 2 2 2 4 6
= (3x − 2ax) + 3 (9x − 12ax + 4a x ) (3a ) + 3 (3x − 2ax) (ga ) + 27a
3
2 2 4 3 3 4 2 4 2 5 6
= (3x − 2ax) + 81a x − 108a x + 36a x + 81a x − 54a x + 27a

3
2 2 4 3 3 4 2 5 6
= (3x − 2ax) + 81a x − 108a x + 117a x − 54a x + 27a

Again by using Binomial Theorem, we obtain
3
2
(3x − 2ax)

3 2
3 2 3 2 3 2 2 3 3
= C0 (3x ) − C1 (3x ) (2ax) + C2 (3x ) (2ax) − C3 (2ax)

6 4 2 2 2 3 3
= 27x − 3 (9x ) (2ax) + 3 (3x ) (4a x ) − 8a x

6 5 2 4 3 3
= 27x − 54ax + 36a x − 8a x
3
2 2
(3x − 2ax + 3a )

6 5 2 4 3 3 2 4 3 3 4 2 5 6
= 27x − 54ax + 36a x − 8a x + 81a x − 108a x + 117a x − 54a x + 27a

6 5 2 4 3 3 4 2 5 6
= 27x − 54ax + 117a x − 116a x + 117a x − 54a x + 27a

Page : 176 , Block Name : Miscellaneous Exercise