Class 11 NCERT Solutions- Chapter 8 Binomial Theorem - Exercise 8.1

The Binomial Theorem provides a method to expand expressions that are raised to a power such as the (x + y)ⁿ. It is a crucial concept in algebra, particularly useful for expanding the polynomials and solving combinatorial problems. In this chapter, students learn to apply the Binomial Theorem to simplify and solve the various algebraic problems.

Binomial Theorem

The Binomial Theorem states that for any positive integer n the expansion of (x + y)ⁿ can be expressed as:

(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k

where \binom{n}{k} is the binomial coefficient, calculated as \frac{n!}{k!(n-k)!}. This theorem allows us to expand expressions involving two terms raised to a power and each term in the expansion is a product of the binomial coefficient and the terms x and y raised to the appropriate powers.

Theorem 1:

(a+b)ⁿ = \sum_{k=0}^{n}  ⁿC_k a^n-k b^k 

Here, the coefficients ⁿC_k are known as binomial coefficients.

Theorem 2:

(a–b)ⁿ = \sum_{k=0}^{n}  (-1)^{n n}C_k a^n-k b^k

Expand each of the expressions in Exercises 1 to 5.

Question 1. (1 – 2x)⁵

Solution:

According to theorem 2, we have

a = 1

b = 2x

and, n = 5

So, (1 - 2x)⁵ = ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (2x)¹ + 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 

Question 2. (\frac{2}{x} – \frac{x}{2})^5

Solution:

According to theorem 2, we have

a = \frac{2}{x}

b = \frac{x}{2}

and, n = 5

So,(\frac{2}{x} – \frac{x}{2})^5  = ⁵C₀ (\frac{2}{x} )⁵ – ⁵C₁ (\frac{2}{x} )⁴ (\frac{x}{2} )¹ + ⁵C₂ (\frac{2}{x} )³ (\frac{x}{2} )² – ⁵C₃ (\frac{2}{x} )² (\frac{x}{2} )³ + ⁵C₄ (\frac{2}{x} )¹ (\frac{x}{2} )⁴ – ⁵C₅ (\frac{x}{2} )⁵

= \frac{32}{x^5}  – 5 (\frac{16}{x^4}) (\frac{x}{2})  + 10 (\frac{8}{x^3}) (\frac{x^2}{4})  – 10 (\frac{4}{x^2})  + 5 (\frac{2}{x}) (\frac{x^4}{16})   – \frac{x^5}{32}

= \frac{32}{x^5} – \frac{40}{x^3} + \frac{20}{x} – 5x + \frac{5x^3}{8} – \frac{x^5}{32}

Question 3. (2x – 3)⁶

Solution:

According to theorem 2, we have

a = 2x

b = 3

and, n = 6

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

Question 4. (\frac{x}{3} + \frac{1}{x})^5

Solution:

According to theorem 1, we have

a = \frac{x}{3}

b = \frac{1}{x}

and, n = 5

So, (\frac{x}{3} + \frac{1}{x})^5  = ⁵C₀ (\frac{x}{3} )⁵ + ⁵C₁ (\frac{x}{3} )⁴ (\frac{1}{x} )¹ + ⁵C₂ (\frac{x}{3} )³ (\frac{1}{x} )² + ⁵C₃ (\frac{x}{3} )² (\frac{1}{x} )³ + ⁵C₄ (\frac{x}{3} )¹ (\frac{1}{x} )⁴ + ⁵C₅ (\frac{1}{x} )⁵

= \frac{x^5}{243} + 5 (\frac{x^4}{81}) (\frac{1}{x}) + 10 (\frac{x^3}{27}) (\frac{1}{x^2}) + 10 (\frac{x^2}{9}) (\frac{1}{x^3}) + 5 (\frac{x}{3}) (\frac{1}{x^4}) + \frac{243}{x^5}

= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}

Question 5. (x + \frac{1}{x})^6

Solution:

According to theorem 1, we have

a = x

b = \frac{1}{x}

and, n = 6

So, (x + \frac{1}{x}) ^ 6  = ⁶C₀ (x)⁶ + ⁶C₁ (x)⁵ (\frac{1}{x} )¹ + ⁶C₂ (x)⁴ (\frac{1}{x} )² + ⁶C₃ (x)³ (\frac{1}{x} )³ + ⁶C₄ (x)² (\frac{1}{x} )⁴ + ⁶C₅ (x)¹ (\frac{1}{x} )⁵ + ⁶C₆ (\frac{1}{x} )⁶

= x^6 + 6(x^5)(\frac{1}{x}) + 15 (x^4) (\frac{1}{x^2}) + 20 (x^3) (\frac{1}{x^3}) + 15 (x^2) (\frac{1}{x^4}) + 6 (x) (\frac{1}{x^5}) + (\frac{1}{x^6})

= x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}

Using the binomial theorem, evaluate each of the following:

Question 6. (96)³ 

Solution:

Given: (96)³

Here, 96 can be expressed as (100 - 4).

So, (96)³ = (100 – 4)³

According to Theorem 2, we have

= 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

Question 7. (102)⁵ 

Solution:

Given: (102)⁵

Here, 102 can be expressed as (100 + 2).

So, here (102)⁵ = (100 + 2)⁵

According to Theorem 1, we have

= 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)³ + 5 (100) (2)⁴ + (2)⁵

= 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

Question 8. (101)⁴

Solution:

Given: (101)⁴

Here, 101 can be expressed as (100 + 1).

So, here (101)⁴ = (100 + 1)⁴

According to Theorem 1, we have

= 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

Question 9. (99)⁵

Solution:

Given: (99)⁵

Here, 99 can be expressed as (100 - 1).

So, here (99)⁵ = (100 – 1)⁵

According to Theorem 2, we have

= 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

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

Solution:

Given: (1.1)¹⁰⁰⁰⁰

Here, 1.1 can be expressed as (1 + 0.1)

So, here (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰

According to Theorem 1, we have

(1 + 0.1)¹⁰⁰⁰⁰ = ¹⁰⁰⁰⁰C₀ (1)¹⁰⁰⁰⁰ + ¹⁰⁰⁰⁰C₁ (1)⁹⁹⁹⁹ (0.1)¹ + other positive terms

= 1 + 1000 + other positive terms

= 1100 + other positive terms

So, 1100 + other positive terms > 1000

Hence, proved (1.1)¹⁰⁰⁰⁰ > 1000

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

Solution:

According to Theorem 1, we have

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

According to Theorem 2, we have

(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)                    -(1)

Now, according to Equation(1), we get

a = √3 and b = √2

So, (√3 + √2)⁴ – (√3 – √2)⁴ 

= 8 × √3 × √2 ((√3)² + (√2)²)

= 8 (√6)(3 + 2)

= 40 √6

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

Solution:

According to Theorem 1, , we have

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

According to Theorem 2, , we have

(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]                           -(1)

Now, According to Equation(1),

x = √2

So, (√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

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

Solution:

To Prove: 9ⁿ⁺¹ – 8n – 9 = 64 k, where k is some natural number

According to Theorem 1, we have

For a = 1, b = 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, will be a natural number

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

Question 14. Prove that \sum_{r=0}^{n}  3^{r n}C_r = 4ⁿ

Solution:

As, we know that According to Binomial Theorem,

\sum_{k=0}^{n}  ⁿC_k a^n-k b^k = (a + b)ⁿ

By comparing Theorem 1 with question, we get

\sum_{r=0}^{n}  3^{r n}C_r = 4ⁿ

a + b = 4, k = r and b = 3

a = 1.

So, \sum_{r=0}^{n}  ⁿC_r a^n-r b^r = (a+b)ⁿ

\sum_{r=0}^{n}  ⁿC_r 1^n-r 3^r = (1+3)ⁿ

\sum_{r=0}^{n}  ⁿC_r (1) 3^r = 4ⁿ

\sum_{r=0}^{n}  ⁿC_r 3_r = 4ⁿ

Hence, Proved 

Read More:

• Binomial Theorem
• Binomial Expansion Formulas