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• NCERT Solutions for Class 11 Maths

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

• Last Updated : 03 Mar, 2021

Theorem 1:

(a+b)n nCk an-k bk

Here, the coefficients nCk are known as binomial coefficients.

Theorem 2:

(a–b)n (-1)n nCk an-k bk

### 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)55C1 (1)4 (2x)1 + 5C2 (1)3 (2x)25C3 (1)2 (2x)3 + 5C4 (1)1 (2x)45C5 (2x)5

= 1 – 5 (2x) + 10 (4x)2 – 10 (8x3) + 5 (16 x4) – (32 x5)

= 1 – 10x + 40x2 – 80x3 + 80x4– 32x5

### Question 2. Solution:

According to theorem 2, we have

a = b = and, n = 5

So, = 5C0 ( )55C1 ( )4 ( )1 + 5C2 ( )3 ( )25C3 ( )2 ( )3 + 5C4 ( )1 ( )45C5 ( )5 – 5 + 10 – 10 + 5 –  ### 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)66C1 (2x)5 (3)1 + 6C2 (2x)4 (3)26C3 (2x)3 (3)3 + 6C4 (2x)2 (3)46C5 (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. Solution:

According to theorem 1, we have

a = b = and, n = 5

So, = 5C0 ( )5 + 5C1 ( )4 ( )1 + 5C2 ( )3 ( )2 + 5C3 ( )2 ( )3 + 5C4 ( )1 ( )4 + 5C5 ( )5  ### Question 5. Solution:

According to theorem 1, we have

a = x

b = and, n = 6

So, = 6C0 (x)6 + 6C1 (x)5 ( )1 + 6C2 (x)4 ( )2 + 6C3 (x)3 ( )3 + 6C4 (x)2 ( )4 + 6C5 (x)1 ( )5 + 6C6 ( )6  ### 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)33C1 (100)2 (4) – 3C2 (100) (4)23C3 (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)55C1 (100)4 (1) + 5C2 (100)3 (1)25C3 (100)2 (1)3 + 5C4 (100) (1)45C5 (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 a44C1 a3 b + 4C2 a2 b24C3 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 a44C1 a3 b + 4C2 a2 b24C3 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 x66C1 x5 + 6C2 x46C3 x3 + 6C4 x26C5 x + 6C6

Now, (x + 1)6 – (x – 1)6

= 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6 – [6C0 x66C1 x5 + 6C2 x46C3 x3 + 6C4 x26C5 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 3rnCr = 4n

Solution:

As, we know that According to Binomial Theorem, nCk an-k bk = (a + b)n

By comparing Theorem 1 with question, we get 3r nCr = 4n

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

a = 1.

So, nCr an-r br = (a+b)n nCr 1n-r 3r = (1+3)n nCr (1) 3r = 4n nCr 3r = 4n

Hence, Proved

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