NCERT Solutions Class 11 Mathematics Chapter 8 Binomial Theorem

Exercise 8.1

1. Expand the expression (1– 2x)⁵

Answer : 

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 

Answer : 

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

3. Expand the expression (2x – 3)⁶ . 

Answer : 

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 

Answer : 

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

5. Expand (x + 1/x)⁶ 

Answer : 

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

6. Using Binomial Theorem, evaluate (96)³

Answer : 

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

Answer : 

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

Answer : 

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

Answer : 

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.

Answer : 

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 

Answer : 

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 

Answer : 

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.

Answer : 

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 + .... + ^mC_m a^m 

For a = 8 and m = n + 1, we obtain 

14. Prove that 

Answer : 

By Binomial Theorem,  

Exercise 8.2

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

Answer : 

It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-rb^r.

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

T_r+1 = ⁸C_r x^8-r 3^r.

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

r = 3 

Thus, the coefficient of x⁵ is ⁸C_r (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 .

Answer : 

It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r .

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 T_r+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 

Answer : 

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

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

T_r+1 = ⁶C_r (x² )^6-r (-y)^r . (-1)^r  6C_r x^12-2r y^r

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

Answer : 

General form of the expansion T_r+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 T_r+1 = ¹²C_r(x²)^12−t(−yx)^r

=(−1)^r12C_r−x^24−2r−y^r

=(−1)^r−2C_rx^24−r⋅y^r

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

Answer : 

It is known that (r + 1)^th term, T_r+1, in the binomial expansion of (a + b)ⁿ is given by T_r+1 =  ⁿC_r a^n-r b^r .

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. 

Answer : 

It is known that (r + 1)th term, (T_r+1 ), in the binomial expansion of (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r.

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 . 

Answer : 

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 = 4^th term and [(7+1)/2 + 1)^th = 5^th term

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

Answer : 

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 a^m and aⁿ are equal.

Answer : 

It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r .

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

T_r+1 =  ^m+nC_r (1)^m+n-r (a)^r . = ^m+nC_r a^r  

Comparing the indices of a in a^m and in T_r+1, we obtain  

r = m 

Therefore, the coefficient of a^m i

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

Answer : 

It is known that (k +1)^th term, (T_k+1 ) in the binomial expansion of (a + b)ⁿ is given by T_k+1 = ⁿC_k a^n-k b^k .

Therefore, (r – 1)^th term in the expansion of (x + 1)ⁿ is T_r-1 = ⁿC_r-2 (x)^n-(r-2) (1)^{(r- 2)} = ⁿC_r-2 X^n-r+2

r^th term in the expansion of (x + 1)ⁿ is T_r = ⁿC_r-1 (x)^n-(r-1) (1)^(r-1) = ⁿC_r-1 x^n-r+1

(r +1)^th term in the expansion of (x +1)ⁿ is T_r+1 = ⁿC_r (X)^n-r (1)^r = ⁿC_r X^n-r

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

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

4r – 12 = 0

⇒ r = 3

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

n – 12 + 5 = 0

⇒ n = 7

Thus, n = 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 .

Answer :  

It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r ..

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 T_r+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

Answer : 

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

T_r+1 =  ⁿC_r a^n-r b^r . 

Assuming that x² occurs in the (r + 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.

Answer : 

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

T_r+1 = ⁿC_r a^n-r b^r .

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

Therefore, we obtain

T₁ = ⁿC₀ a^n-0 b⁰ = aⁿ = 729  ...(1)

T₂ = ⁿC₁ a^n-1 b¹ = na^n-1 b = 7290  ...(2)

T₃ = ⁿC₂ a^n-2 b² = [n(n-1)/2] a^n-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.

Answer : 

It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r ..

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.

Answer : 

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]

Answer : 

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

aⁿ – bⁿ = (a – b) [(a –b)^n-1 + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n bⁿ]

aⁿ – bⁿ = (a – b) ^k

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

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

Answer : 

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

Answer : 

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.

Answer : 

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 

Answer : 

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b + ⁿC₂a^n-2b² + ... + ⁿC_n-1 ab^n-1 + ⁿC_nbⁿ,

Fifth term from the beginning = ⁿC₄ a^n-4 b⁴

Fifth term from the end = ⁿC_n-4 a⁴ b^n-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 

Answer : 

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.

Answer : 

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

= 2^7x6 – 54ax⁵+ 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Thus, (3x² – 2ax + 3a²)³

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