# NCERT Solutions Class 11 Mathematics Chapter 8 Binomial Theorem

## Chapter 8 Binomial Theorem Class 11 Mathematics NCERT Solutions

Class 11 Mathematics students should refer to the following NCERT questions with answers for Chapter 8 Binomial Theorem in Class 11. These NCERT Solutions with answers for Class 11 Mathematics will come in exams and help you to score good marks

### Chapter 8 Binomial Theorem NCERT Solutions Class 11 Mathematics

Exercise 8.1

1. Expand the expression (1– 2x)5

By using Binomial Theorem, the expression (1-2x) can be expanded as

(1 -2x)5

5C0 (1)5 – 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 - 5C3 (1)2 (2x)3 + 5C4 (1)1(2x)4 + 5C5 (2x)5

= 1 – 5(2x) + 10(4x2 ) – 10(8x3 )+ 5(16x4 ) – (32x5 )

= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5

2. Expand the expression (2/x - x/2)5

By using Binomial Theorem, the expression (2/x - x/2)5 can be expanded as

3. Expand the expression (2x – 3)6 .

By using Binomial theorem, the expression (2x - 3)6 can be expanded as

(2x - 3)6

6C0 (2x)6 – 6C1 (2x)5 (3) + 6C2 (2x)4 (3)2 – 6C3 (2x)3 (3)3 + 6C4 (2x)2 (3)4 – 6C5 (2x) (3)5 + 6C6 (3)6

= 64x6 – 6(32x5 )(3) + 15(16x4 )(9) – 20(8x3 )(27) + 15(4x2) (81) – 6(2x)(243) + 729

= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

4. Expand the expression (x/3 + 1/x)5

By using Binomial Theorem, the expression (x/3 + 1/x)5 can be expanded as

5. Expand (x + 1/x)6

By using Binomial Theorem, the expression(x + 1/x)6 can be expanded as

6. 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

∴ (96)3 = (100 - 4)3

3C0 (100)3 - 3C1 (100)2 (4) + 3C2 (100)(4)2 - 3C3 (4)3

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

= 1000000 - 120000 + 4800 - 64

= 884736

7. 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

∴ (102)5 = (100 + 2)5

5C0(100)5 - 5C1(100)4 (2) + 5C2(100)3 (2)2 - 5C3(100)2 (2)3 + 5C4(100)(2)4 + 5C5(2)5

= (100)5 + 5(100)4 (2) + 10(100)3 (2)2 + 10(100)2 (2)3 + 5(100)(2)4 +(2)5

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

8. 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

∴ (101)4  = (100 + 1)4

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

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

= 100000000 +4000000 + 60000 + 400 + 1

= 104060401

9. 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

∴ (99)5 = (100 - 1)5

5C0 (100)5 - 5C1 (100)4 (1) + 5C2 (100)3 (1)2 - 5C3 (100)2 (1)3 + 5C4 (100)(1)4 + 5C5 (1)5

= (100)5 - 5(100)4 + 10(100)3 - 10(100)2 + 5(100) - 1

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

= 10010000500 - 500100001

= 9509900499

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

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

(1.1)10000 = (1 + 0.1)10000

10000C0 - 10000C1 (1.1) + Other positive terms

= 1 + 10000 × 1.1 + Other positive terms

= 1 + 11000 + Other positive terms

> 1000

Hence, (1.1)10000 > 1000

11. Find (a + b)4 - (a - b)4. Hence, evaluate (√3 + √2)4 - (√3 - √2)4

Using Binomial Theorem, the expressions, (a + b)4 and (a - b)4 , can be expanded as

(a + b)4 = 4C0a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4

(a - b)4 =  4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4

∴ (a + b)4 - (a - b)4 = 4C0a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4 - [4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4]

= 2(4C1a3b + 4C3ab3) = 2(4a3b + 4ab3

= 8ab (a2 + b2 )

By putting a = √3 and b = √2 , we obtain

(√3 + √2)4 - (√3 - √2)4 = 8(√3)(√2) {(√3)2  + (√2)2 }

= 8(√6) (3 + 2) = 40√6

12. Find (x + 1)6 + (x -1)6. Hence or otherwise evaluate (√2 + 1)6 + (√2 - 1)6

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

(x + 1)6  = 6C0 x6 + 6C1 x5  + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6

(x - 1)6 =  6C0 x6 - 6C1 x5  + 6C2 x4 - 6C3 x3 + 6C4 x2 - 6C5 x + 6C6

∴ (x + 1)6 + (x -1)6 = 2[6C0 x6 + 6C1 x5  + 6C2 x4 +  6C4 x2 + 6C6]

= 2[x6 + 15x4 + 15x2 + 1]

By putting x = √2, we obtain

(√2 + 1)6  + (√2 - 1)6  = 2[(√2)6 + 15(√2)4 + 15(√2)2 + 1 ]

= 2(8 + 15 × 4 + 15 × 2 + 1)

= 2(8 + 60 + 30 + 1)

= 2(99) = 198

13. Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

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

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

By Binomial Theorem,

(1 + a)m =  mC0 + mC1 a + mC2 a2 + .... + mCm am

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

14. Prove that

By Binomial Theorem,

Exercise 8.2

1. Find the coefficient of x5 in (x + 3)8

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-rbr.

Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8 , we obtain

Tr+1 = 8Cr x8-r 3r.

Comparing the indices of x in x5 and in Tr+1, we obtain

r = 3

Thus, the coefficient of x5 is 8Cr (3)3 = (8!/3!5!) × 33 = (8.7.6.5!/3.2.5!).33 = 1512

2. Find the coefficient of a5 b7 in (a - 2b)12 .

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by Tr+1 = nCr an-r br .

Assuming that a5 b7 occurs in the (r + 1)th term of the expansion (a - 2b)12 , we obtain

Comparing the indices of a and b in a5 b7 and in Tr+1 , we obtain
r = 7
Thus, the coefficient of a5 b7 is 12C7 (-2)7 = (-12!/7!5!) .27 = (12.11.10.9.8.7!/5.4.3.2.7!) 27 = -(792)(128) = -101376

3. Write the general term in the expansion of (x2 - y)6

It is known that the general term Tr+1{which is the (r+1)th term} in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br .

Thus, the general term in the expansion of (x2 - y6) is

Tr+1 = 6Cr (x2 )6-r (-y)r . (-1)r  6Cr x12-2r yr

Question4. Write the general term in the expansion of (x2 – yx)12, x ≠ 0

General form of the expansion Tr+1 is

{ which is the (r+1)th term } in the binomial expansion of (a+b)n is given by Tr+1=nCran−rbr.

Thus, the general term in the expansion of (x2−yx)12

is Tr+1 12Cr(x2)12−t(−yx)r

=(−1)r12Cr−x24−2r−yr

=(−1)r−2Crx24−r⋅yr

5. Find the 4th term in the expansion of (x – 2y)12 .

It is known that (r + 1)th term, Tr+1, in the binomial expansion of (a + b)n is given by Tr+1 =  nCr an-r br .

Thus, the 4th term in the expansion of (x - 2y)12 is

6. Find the 13th term in the expansion of (9x - 1/3√x)18 , x ≠ 0.

It is known that (r + 1)th term, (Tr+1 ), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br.

Thus, 13th term in the expansion of (9x - 1/3√x)18 is

7. Find the middle terms in the expansions of (3 - x3 /6)7 .

It is known that in the expansion of (a + b)n, if n is odd, then there are two middle terms, namely, [(n+1)/2]8 term and [(n+1)/2 + 1)th term.

Therefore, the middle terms in the expansion of [3 - x3 /6]7 are [(7+1)/2]th = 4th term and [(7+1)/2 + 1)th = 5th term

8. Find the middle terms in the expansions of (x/3 + 9y)10

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

Therefore, the middle term in the expansion of (x/3 + 9y)10 is (10/2 + 1)th = 6th term

Thus, the middle term in the expansion of (x/3 + 9y)10 is 61236 x5 y5.

9. In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by Tr+1 = nCr an-r br .

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

Tr+1 =  m+nCr (1)m+n-r (a)r . = m+nCr ar

Comparing the indices of a in am and in Tr+1, we obtain

r = m

Therefore, the coefficient of am i

10. The coefficients of the (r – 1)thrth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1:3:5. Find n and r.

It is known that (k +1)th term, (Tk+1 ) in the binomial expansion of (a + b)n is given by Tk+1 = nCk an-k bk .

Therefore, (r – 1)th term in the expansion of (x + 1)n is Tr-1 = nCr-2 (x)n-(r-2) (1)(r- 2) = nCr-2 Xn-r+2

rth term in the expansion of (x + 1)n is Tr = nCr-1 (x)n-(r-1) (1)(r-1) = nCr-1 xn-r+1

(r +1)th term in the expansion of (x +1)n is Tr+1 = nCr (X)n-r (1)r = nCr Xn-r

Therefore, the coefficients of the (r -1)th , rth and (r + 1)th terms in the expansion of (x + 1)n are nCr-2 , nCr-1 and nCr respectively. Since these coefficients are in the ratio 1 : 3 : 5, we obtain

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

4– 12 = 0

⇒ r = 3

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

n – 12 + 5 = 0

⇒ n = 7

Thus, = 7 and r = 3

11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by Tr+1 = nCr an-r br ..

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

Comparing the indices of x in xn and in Tr+1 , we obtain

r = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n is

Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.

Hence, proved.

12. Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6

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

Tr+1 =  nCr an-r br .

Assuming that x2 occurs in the (+ 1)th term of the expansion (1 +x)m, we obtain

⇒ m(m - 1) = 12

⇒ m2 - m - 12 = 0

⇒ m2 - 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

Thus, the positive value of m, for which the coefficient of x2 in the expansion

(1 + x)m is 6, is 4.

Miscellaneous Solutions

1. Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.

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

Tr+1 = nCr an-r br .

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

Therefore, we obtain

T1 = nC0 an-0 b0 = an = 729  ...(1)

T2 = nC1 an-1 b1 = nan-1 b = 7290  ...(2)

T3 = nC2 an-2 b= [n(n-1)/2] an-2 b2  = 30375 ...(3)

Dividing (2) by (1), we obtain

⇒ n = 6

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

a6 = 729

⇒ a = 6√729 = 3

From (5), we obtain

b/3 = 5/3 ⇒ 5

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

2. Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by Tr+1 = nCr an-r br ..

Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain

It is given that the coefficients of x2 and x3 are the same.

84(3)6 a3 = 36(3)7 a2

⇒ 84a = 36× 3

⇒ a = (36× 3)/84 = 104/84

⇒ a = 9/7

Thus, the required value of a is 9/7.

3. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Question4. If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.

[Hint: write an = (a – b + b)n and expand]

n order to prove that (a – b) is a factor of (an – bn), it has to be proved that

an – bn = k (a – b) where k is some natural number.

a can be written as a = a – b + b

an = (a – b + b)n = [(a – b) + b]n

nC0 (a – b)n + nC1 (a – b)n-1 b + …… + n C n bn

a– bn = (a – b) [(a –b)n-1 + nC1 (a – b)n-1 b + …… + bn]

an – bn = (a – b) k

Where k = [(a –b)n-1 nC1 (a – b)n-1 b + …… + n C n bn] is a natural number

Question5. Evaluate: (√3 + √2)- (√3 - √2)6

6. Find the value of [a2 + √(a2 - 1)]4 + [a2 - √(a2 - 1)]4

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

This can be done as

7. Find an approximation of (0.99)5 using the first three terms of its expansion.

0.99 = 1 - 0.01

∴ (0.99)5 = (1 - 0.01)5

5C0 (1)5 - 5C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2  (Approximately)

= 1 - 5(0.01) + 10(0.01)2

1 - 0.05 + 0.001

= 1.001 - 0.05

= 0.951

Thus, the value of (0.99)5 is approximately 0.951.

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

(a + b)n = nC0an + nC1an-1b + nC2an-2b2 + ... + nCn-1 abn-1 + nCnbn,

Fifth term from the beginning = nC4 an-4 b4

Fifth term from the end = nCn-4 a4 bn-4

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is √6 : 1. Therefore, from (1) and (2), we obtain

9. Expand using Binomial Theorem (1 + x/2 - 2/x)4, x ≠ 0

Using Binomial Theorem, the given expression (1 + x/2 - 2/x)4 can be expanded as

Question10. Find the expansion of (3x2 – 2ax + 3a2)3using binomial theorem.

Here

=  We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3

Putting a = 3x2 & b = -a (2x-3a), we get

[3x2 + (-a (2x-3a))]3

= (3x2)3+3(3x2)2(-a (2x-3a)) + 3(3x2) (-a (2x-3a))2 + (-a (2x-3a))3

= 27x6 – 27ax4 (2x-3a) + 9a2x2 (2x-3a)2 – a3(2x-3a)3

= 27x6 – 54ax5 + 81a2x4 + 9a2x2 (4x2-12ax+9a2) – a3 [(2x)3 – (3a)3 – 3(2x)2(3a) + 3(2x)(3a)2]

= 27x6 – 54ax5 + 81a2x4 + 36a2x4 – 108a3x3 + 81a4x2 – 8a3x3 + 27a6 + 36a4x2 – 54a5x

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6

Thus, (3x2 – 2ax + 3a2)3

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6

 NCERT Solutions Class 11 Mathematics Chapter 1 Sets
 NCERT Solutions Class 11 Mathematics Chapter 2 Relations and Functions
 NCERT Solutions Class 11 Mathematics Chapter 3 Trigonometric Functions
 NCERT Solutions Class 11 Mathematics Chapter 4 Principle of Mathematical Induction
 NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations
 NCERT Solutions Class 11 Mathematics Chapter 6 Linear Inequalities
 NCERT Solutions Class 11 Mathematics Chapter 7 Permutations and Combinations
 NCERT Solutions Class 11 Mathematics Chapter 8 Binomial Theorem
 NCERT Solutions Class 11 Mathematics Chapter 9 Sequences and Series
 NCERT Solutions Class 11 Mathematics Chapter 10 Straight Lines
 NCERT Solutions Class 11 Mathematics Chapter 11 Conic Sections
 NCERT Solutions Class 11 Mathematics Chapter 12 Introduction to Three Dimensional Geometry
 NCERT Solutions Class 11 Mathematics Chapter 13 Limits and Derivatives
 NCERT Solutions Class 11 Mathematics Chapter 14 Mathematical Reasoning
 NCERT Solutions Class 11 Mathematics Chapter 15 Statistics
 NCERT Solutions Class 11 Mathematics Chapter 16 Probability

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