Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8

Question 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.

Solutions:

As, we know that (r+1)^th term of (a+b)ⁿ is denoted by,

T_r+1 = ⁿC_r a^n-r b^r

Here, it is given that first three terms of the expansion are 729, 7290 and 30375.

When, T₁ = 729, T₂ = 7290 and T₃ = 30375

T₀₊₁ (r=0) = ⁿC₀ a^n-0 b⁰ = aⁿ = 729  …………………(1)

T₁₊₁ (r=0) = ⁿC₁ a^n-1 b¹ = na^n-1 b = 7290  …………………(2)

T₂₊₁ (r=0) = ⁿC₂ a^n-2 b² = a^n-2b² =  = 30375  …………………(3)

Dividing (1) and (2), we get

na^n-1-n b = 10

 = 10 ……………………….(I)

Now, dividing (3) and (2), we get

From (I), we can substitute

 = 10 – 

 ………………….(II)

Substituting (II) in (I), we get

 = 10

 = 10

n = 

n = 6

Substituting n = 6 in (1), we get

aⁿ = 729

a⁶ = 729

a = 3

Substituting a = 3 in (II), we get

b = 

b = 5

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

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

Solutions:

As, we know that (r+1)^th term of (a+b)ⁿ is denoted by,

T_r+1 = ⁿC_r a^n-r b^r

Here, a = 3 and b = ax and n = 9.

T_r+1 = ⁹C_r 3^9-r (ax)^r

T_r+1 = ⁹C_r  a^rx^r

T_r+1 = ⁹C_r 

So, here if you want the power of x² and x³. Then r=2 and r=3.

r = 2, T_{2+1 =} 

r = 3, T₃₊₁ = 

Coefficient of x² = Coefficient of x³

a = 

Hence, a = 

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

Solutions:

For getting the coefficient of x⁵, lets expand both the binomials for more clear understanding.

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

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

= 1 + 12 x + 60x² + 160 x³ + 240 x⁴ + 192 x⁵ + 64x⁶

(1 – x)⁷ = ⁷C₀ – ⁷C₁ (x) + ⁷C₂ (x)² – ⁷C₃ (x)³ + ⁷C₄ (x)⁴ – ⁷C₅ (x)⁵ + ⁷C₆ (x)⁶ – ⁷C₇ (x)⁷

= 1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + 7x⁶ – x⁷

Now, the product will be seen as follows

(1 + 2x)⁶ (1 – x)⁷ = (1 + 12 x + 60x² + 160 x³ + 240 x⁴ + 192 x⁵ + 64x⁶) (1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + 7x⁶ – x⁷)

Here, what we can see that x⁵ will be obtained when two terms will be multiplied of having sum of power of x as 5. Those two terms will be as follows:

First binomial	Second binomial
1st term	6th term
2nd term	5th term
3rd term	4th term
4th term	3rd term
5th term	2nd term
6th term	1st term

So, 

Coefficient of x⁵ = (1)(-21) + (12)(35) + (60)(-35) + (160)(21) + (240)(-7) + (192)(1)

Coefficient of x^{5 =} -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x⁵ = -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x⁵ = 171

Hence, the coefficient of x⁵ in the expression (1+2x)⁶ (1-x)⁷ is 171.

Question 4. 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]

Solutions:

To prove that (a – b) is a factor of (aⁿ – bⁿ), 

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

a can be written as = a – b + b

aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ

[(a – b) + b]ⁿ = ⁿC₀ (a – b)ⁿ + ⁿC₁ (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + ⁿC_n bⁿ

aⁿ = (a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + bⁿ

Now, aⁿ – bⁿ will be

aⁿ – bⁿ = [(a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + bⁿ] – bⁿ

aⁿ – bⁿ = (a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1

Taking (a-b) common, we have

aⁿ – bⁿ = (a – b) [(a –b)^n-1 + n (a – b)^n-2 b + …… + ⁿC_n-1 b^n-1]

aⁿ – bⁿ = (a – b) k

Where k = [(a –b)^n-1 + n (a – b)^n-2 b + …… + ⁿC_n-1 b^n-1] is a natural number

Hence, it is proved a – b is a factor of aⁿ – bⁿ, where n is a positive integer

Question 5. Evaluate (√3​+√2​)⁶−(√3​−√2​)⁶.

Solutions:

Using binomial theorem the expression (a + b)⁶ and (a – b)⁶, can be expanded as follows:

(a + b)⁶ = ⁶C0 a⁶ + ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² + ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ + ⁶C₅ a b⁵ + ⁶C₆ b⁶

(a – b)⁶ = ⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵ + ⁶C₆ b⁶

Now adding them,

(a + b)⁶ – (a – b)⁶ = ⁶C₀ a⁶ + ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² + ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ + ⁶C₅ a b⁵ + ⁶C₆ b⁶ – [⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵ + ⁶C₆ b⁶]

(a + b)⁶ – (a – b)⁶ = 2[⁶C₁ a⁵ b + ⁶C₃ a³ b³ + ⁶C₅ a b⁵]

Substituting a = √3 and b = √2, we get

(√3 + √2)⁶ – (√3 – √2)⁶ = 2 [6 (√3)⁵ (√2) + 20 (√3)³ (√2)³ + 6 (√3) (√2)⁵]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

Question 6. Find the value of .

Solutions:

Using binomial theorem the expression (x+y)⁴ and (x – y)⁴, can be expanded as follows:

(x + y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴

(x – y)⁴ = ⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄  y⁴

Now adding them,

(x + y)⁴ + (x – y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴ + [⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄ y⁴]

(x + y)⁴ + (x – y)⁴ = 2[⁴C₀ x⁴ + ⁴C₂ x² y² + ⁴C₄ y⁴]

Substituting x = a² and y = , we get

= 2[a⁸ + 6a⁴ (a²-1) + (a²-1)²]

= 2[a⁸ + 6a⁶ – 6a⁴ + (a⁴ + 1 – 2(a²)(1))]

= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]

= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

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

Solutions:

To make 0.99 in binomial form,

0.99 = 1 – 0.01

Now by applying binomial theorem, we get

(0. 99)⁵ = (1 – 0.01)⁵

Taking first three terms of its expansion, we have

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

= 1 – 5 (0.01) + 10 (0.01)²

= 1 – 0.05 + 0.001

= 0.951

Approximation of (0.99)⁵ = 0.951.

Question 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of  is √6:1.

Solutions:

As, here it is said we have to calculate

(Fifth term from the beginning : Fifth term from the end) of the Binomial = 

Instead of taking fifth term from the end, lets reverse the binomial term and take it from beginning.

Fifth term from the end of binomial  = Fifth term from the beginning 

Lets move further with this

As, we know that (r+1)^th term of (a+b)ⁿ is denoted by,

T_r+1 = ⁿCr a^n-r b^r

Fifth term from the beginning of , 

T₅ = T₄₊₁ = ⁿC₄ 

T₅ = ⁿC₄ 

T₅ = ⁿC⁴ 

T₅ = ⁿC₄  …………………………….(1)

Now, Fifth term from the beginning of ,

T₅ = T₄₊₁ = ⁿC₄ 

T₅ = ⁿC₄ 

T₅ = ⁿC₄ 

T₅ = ⁿC₄ 

T₅ = ⁿC₄  ……………………….(2)

Now taking ratio of (1) and (2), which is equal to √6:1

n = 

n = 10

Question 9. Expand using Binomial Theorem , x≠0.

Solutions:

Grouping  in binomial form, we have

Comparing it with (a+b)ⁿ,

a = , b =  and n = 4

 = ⁴C₀  –  ⁴C₁  + ⁴C₂  – ⁴C₃  + ⁴C₄ 

= 

= 

= 

Now, lets get the value of  and 

 = ³C₀ (1)³ + ³C₁ (1)²  + ³C₂ (1)  + ³C₃ 

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

Now, substituting these values in the main equation, we get

= 

= 

= 

= 

Question 10. Find the expansion of (3x²– 2ax + 3a²)³ using binomial theorem.

Solutions:

Grouping (3x²– 2ax + 3a²)³ in binomial form, we have

[3x² + (- 2ax + 3a²)]³

Comparing it with (a+b)ⁿ,

a = 3x², b = -a (2x-3a) and n = 3

[3x² + (-a (2x-3a))]³

= ³C₀ (3x²)³ +  ³C₁ (3x²)² (-a (2x-3a)) +  ³C₂ (3x²) (-a (2x-3a))² +  ³C₃ (-a (2x-3a))³

= 27x⁶ +  3 (9x⁴) (-a) (2x-3a) +  3 (3x²) (-a)² (2x-3a)² + (-a)³ (2x-3a)³

= 27x⁶ + (-54ax⁵ + 81a²x⁴) +  9a²x² (2x-3a)² – a³ (2x-3a)³

Now, lets get the value of (2x-3a)² and (2x-3a)³.

(2x-3a)² = (2x)² + (3a)² – 2(2x)(3a)

(2x-3a)² = 4x² + 9a² -12xa

(2x-3a)³ = (2x)³ – (3a)³ – 3(2x)(3a)(2x-3a)

(2x-3a)³ = 8x³ – 27a³ – 36x²a +54xa²

Now, substituting these values in the main equation, we get

= 27x⁶ – 54ax⁵ + 81a²x⁴ +  9a²x² (4x² + 9a² -12xa) – a³ (8x³ – 27a³ – 36x²a + 54xa²)

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 36a²x⁴ + 81a⁴x² -108x³a³ – (8a³x³ – 27a⁶ – 36x²a⁴ + 54xa⁵⁾

= 27x⁶ – 54ax⁵ + 117a²x⁴ + 81a⁴x² -108x³a³ – 8a³x³ + 27a⁶ + 36x²a⁴ – 54xa⁵

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