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

The NCERT solutions of the first exercise of Class 11 Chapter 8 are available here. These solutions are present in the  PDF format to help students with their studies. The Exercise 8.1 of NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem is based on the following topics:

1. Introduction to Binomial Theorem
2. Binomial Theorem for Positive Integral Indices
3. Pascal’s Triangle
• Binomial theorem for any positive integer n,
• Some special cases

NCERT textbook contains numerous questions in it, which are intended for the students to solve and practice. To score high marks in the Class 11 examination, solving and practicing the NCERT Solutions for Class 11 Maths is a must.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem Exercise 8.1      ### Solutions for Class 11 Maths Chapter 8 – 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

= 5Co (1)55C1 (1)4 (2x) + 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 – 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)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

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)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

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 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)

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 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]

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 am

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: ### Access Other Exercise Solutions of Class 11 Maths Chapter 8- Binomial Theorem

Exercise 8.2 Solutions 12 Questions

Miscellaneous Exercise On Chapter 8 Solutions 10 Questions