NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

NCERT Class 11 Maths Chapter 8

Exercise 8.1 Page No: 166

Expand each of the expressions in Exercises 1 to 5.

1. (1 – 2x)⁵

Solution:

From binomial theorem expansion we can write as

(1 – 2x)⁵

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

= 1 – 10x + 40x² – 80x³ – 32x⁵

Solution:

From binomial theorem, given equation can be expanded as

3. (2x – 3)⁶

Solution:

From binomial theorem, given equation can be expanded as

Solution:

From binomial theorem, given equation can be expanded as

Solution:

From binomial theorem, given equation can be expanded as

6. (96)³

Solution:

Given (96)³

96 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 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. (102)⁵

Solution:

Given (102)⁵

102 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 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 + 5 (100) (2)³ + 5 (100) (2)⁴ + (2)⁵

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

= 11040808032

8. (101)⁴

Solution:

Given (101)⁴

101 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 101 = 100 + 1

(101)⁴ = (100 + 1)⁴

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

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

= 100000000 + 400000 + 60000 + 400 + 1

= 1040604001

9. (99)⁵

Solution:

Given (99)⁵

99 can be written as the sum or difference of two numbers then binomial theorem can be applied.

The given question can be written as 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

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

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

Solution:

By splitting the given 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

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

= 1 + 10000 × 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)¹⁰⁰⁰⁰ > 1000

11. Find (a + b)⁴ – (a – b)⁴. Hence, evaluate 

Solution:

Using binomial theorem the expression (a + b)⁴ and (a – b)⁴, can be expanded

(a + b)⁴ = ⁴C₀ a⁴ + ⁴C₁ a³ b + ⁴C₂ a² b² + ⁴C₃ a b³ + ⁴C₄ b⁴

(a – b)⁴ = ⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b² – ⁴C₃ a b³ + ⁴C₄ b⁴

Now (a + b)⁴ – (a – b)⁴ = ⁴C₀ a⁴ + ⁴C₁ a³ b + ⁴C₂ a² b² + ⁴C₃ a b³ + ⁴C₄ b⁴ – [⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b² – ⁴C₃ a b³ + ⁴C₄ b⁴]

= 2 (⁴C₁ a³ b + ⁴C₃ a b³)

= 2 (4a³ b + 4ab³)

= 8ab (a² + b²)

Now by substituting a = √3 and b = √2 we get

(√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 

 

Solution:

Using binomial theorem the expressions, (x + 1)⁶ and (x – 1)⁶ can be expressed 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

Now, (x + 1)⁶ – (x – 1)⁶ = ⁶C₀ x⁶ + ⁶C₁ x⁵ + ⁶C₂ x⁴ + ⁶C₃ x³ + ⁶C₄ x² + ⁶C₅ x + ⁶C₆ – [⁶C₀ x⁶ – ⁶C₁ x⁵ + ⁶C₂ x⁴ – ⁶C₃ x³ + ⁶C₄ x² – ⁶C₅ x + ⁶C₆]

= 2 [⁶C₀ x⁶ + ⁶C₂ x⁴ + ⁶C₄ x² + ⁶C₆]

= 2 [x⁶ + 15x⁴ + 15x² + 1]

Now by substituting x = √2 we get

(√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.

Solution:

In order to show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, it has to be show that 9ⁿ⁺¹ – 8n – 9 = 64 k, where k is some natural number

Using binomial theorem,

(1 + a)^m = ^mC₀ + ^mC₁ a + ^mC₂ a² + …. + ^m C _m a^m

For a = 8 and m = n + 1 we get

(1 + 8)ⁿ⁺¹ = ⁿ⁺¹C₀ + ⁿ⁺¹C₁ (8) + ⁿ⁺¹C₂ (8)² + …. + ⁿ⁺¹ C _n+1 (8)ⁿ⁺¹

9ⁿ⁺¹ = 1 + (n + 1) 8 + 8² [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)^n-1]

9ⁿ⁺¹ = 9 + 8n + 64 [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)^n-1]

9ⁿ⁺¹ – 8n – 9 = 64 k

Where k = [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)^n-1] is a natural number

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

Hence the proof

14. Prove that 

Solution:

---

Exercise 8.2 Page No: 171

Find the coefficient of

1. x⁵ in (x + 3)⁸

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here x⁵ is the T_r+1 term so a= x, b = 3 and n =8

T_r+1 = ⁸C_r x^8-r 3^r…………… (i)

For finding out x⁵

We have to equate x⁵= x^8-r

⇒ r= 3

Putting value of r in (i) we get

= 1512 x⁵

Hence the coefficient of x⁵= 1512

2. a⁵b⁷ in (a – 2b)¹² .

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here a = a, b = -2b & n =12

Substituting the values, we get

T_r+1 = ¹²C_r a^12-r (-2b)^r………. (i)

To find a⁵

We equate a^12-r =a⁵

r = 7

Putting r = 7 in (i)

T₈ = ¹²C₇ a⁵ (-2b)⁷

= -101376 a⁵ b⁷

Hence the coefficient of a⁵b⁷= -101376

Write the general term in the expansion of

3. (x² – y)⁶

Solution:

The general term T_r+1 in the binomial expansion is given by

T_r+1 = ⁿ C _r a^n-r b^r…….. (i)

Here a = x² , n = 6 and b = -y

Putting values in (i)

T_r+1 = ⁶C_r x ^2(6-r) (-1)^r y^r

= -1^{r 6}c_r .x^{12 – 2r}. y^r

4. (x² – y x)¹², x ≠ 0.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here n = 12, a= x² and b = -y x

Substituting the values we get

T_n+1 =¹²C_r × x^2(12-r) (-1)^r y^r x^r

= -1^{r 12}c_r .x^{24 –2r}. y^r

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

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here a= x, n =12, r= 3 and b = -2y

By substituting the values we get

T₄ = ¹²C₃ x⁹ (-2y)³

= -1760 x⁹ y³

6. Find the 13^th term in the expansion of 

Solution:

Find the middle terms in the expansions of

Solution:

Solution:

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

Solution:

We know that the general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

T_r+1 = ^m+n C_r 1^m+n-r a^r

= ^m+n C_r a^r…………. (i)

Now we have that the general term for the expression is,

T_r+1 =  ^m+n C_r a^r

Now, For coefficient of a^m

T_m+1 =  ^m+n C_m a^m

Hence, for coefficient of a^m, value of r = m

So, the coefficient is ^m+n C _m

Similarly, Coefficient of aⁿ is ^m+n C _n

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.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here the binomial is (1+x)ⁿ with a = 1 , b = x and n = n

The (r+1)^th term is given by

T_(r+1) = ⁿC_r 1^n-r x^r

T_(r+1) = ⁿC_r x^r

The coefficient of (r+1)^th term is ⁿC_r

The r^th term is given by (r-1)^th term

T_(r+1-1) = ⁿC_r-1 x^r-1

T_r = ⁿC_r-1 x^r-1

∴ the coefficient of r^th term is ⁿC_r-1

For (r-1)^th term we will take (r-2)^th term

T_r-2+1 = ⁿC_r-2 x^r-2

T_r-1 = ⁿC_r-2 x^r-2

∴ the coefficient of (r-1)^th term is ⁿC_r-2

Given that the coefficient of (r-1)^th, r^th and r+1^th term are in ratio 1:3:5

Therefore,

⇒ 5r = 3n – 3r + 3

⇒ 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equation 2 and 3

2n -8r +10 =0

-3n – 8r – 3 =0

⇒ -n = -7

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}.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

The general term for binomial (1+x)²ⁿ is

T_r+1 = ²ⁿC_r x^r …………………..1

To find the coefficient of xⁿ

r = n

T_n+1 = ²ⁿC_n xⁿ

The coefficient of xⁿ = ²ⁿC_n

The general term for binomial (1+x)^2n-1 is

T_r+1 = ^2n-1C_r x^r

To find the coefficient of xⁿ

Putting n = r

T_r+1 = ^2n-1C_r xⁿ

The coefficient of xⁿ = ^2n-1C_n

We have to prove

Coefficient of xⁿ in (1+x)²ⁿ = 2 coefficient of xⁿ in (1+x)^2n-1

Consider LHS = ²ⁿC_n

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

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a = 1, b = x and n = m

Putting the value

T_r+1 = ^m C_r 1^m-r x^r

= ^m C_r x^r

We need coefficient of x²

∴ putting r = 2

T₂₊₁ = ^mC₂ x²

The coefficient of x² = ^mC₂

Given that coefficient of x² = ^mC₂ = 6

⇒ m (m – 1) = 12

⇒ m²– m – 12 =0

⇒ m²– 4m + 3m – 12 =0

⇒ m (m – 4) + 3 (m – 4) = 0

⇒ (m+3) (m – 4) = 0

⇒ m = – 3, 4

We need positive value of m so m = 4

---

Miscellaneous Exercise Page No: 175

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.

Solution:

We know 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-t b^r

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

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 get

\frac{na^{n-1}b}{a^{n}} = \frac{7290}{729} \\ \frac{nb}{a} = 10annan−1b​=7297290​anb​=10

Dividing 3 by 2 we get

\frac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \frac{30375}{7290} \\ \frac{(n-1)b}{2a} = \frac{30375}{7290} \\ \frac{(n-1)b}{a} = \frac{30375}{7290} \times 2 =\frac{25}{3} \\ \frac{(nb)}{a} -\frac{b}{a} = \frac{25}{3} \\ 10 -\frac{b}{a} = \frac{25}{3} \\ \frac{b}{a} = 10 – \frac{25}{3} = \frac{5}{3}2nan−1bn(n−1)an−2b2​=729030375​2a(n−1)b​=729030375​a(n−1)b​=729030375​×2=325​a(nb)​−ab​=325​10−ab​=325​ab​=10–325​=35​

From 4 and 5 we have

n. 5/3 = 10

n = 6

Substituting n = 6 in 1 we get

a⁶ = 729

a = 3

From 5 we have, b/3 = 5/3

b = 5

Thus a = 3, b = 5 and n = 76

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

Solution:

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

Solution:

(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

(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)

192 – 21 = 171

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

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]

Solution:

In 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 bⁿ] is a natural number

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

5. Evaluate 

Solution:

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

(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 (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]

Now by 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

6. Find the value of 

Solution:

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

Solution:

0.99 can be written as

0.99 = 1 – 0.01

Now by applying binomial theorem we get

(o. 99)⁵ = (1 – 0.01)⁵

= 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

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

Solution:

9. Expand using Binomial Theorem 

Solution:

Using binomial theorem the given expression can be expanded as

Again by using binomial theorem to expand the above terms we get

From equation 1, 2 and 3 we get

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

Solution:

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