Exercise 8.1

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 + 80x⁴– 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)³ + ⁴C₄ (1)⁴

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

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

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 – ⁶C₅x + 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

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