# Binomial Theorem Chapter 8 Class 11 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 11 Maths Chapter 8‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Maths Chapter 8 Exercise Solution’ using the download button.

### Chapter 8: Binomial Theorem

#### 8.1 Introduction

In earlier classes, we have learned how to find the squares and cubes of binomials like a + b and a – b. Using them, we could evaluate the numerical values of numbers like (98)2 = (100 – 2)2 , (999)3 = (1000 – 1) 3 , etc. However, for higher powers like (98)5, (101)6, etc.,

the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as the binomial theorem.

It gives an easier way to expand (a + b)n, where n is an integer or a rational number. In this chapter, we study the binomial theorem for positive integral indices only.

#### Pascal’s Triangle

The structure given in Fig 8.2 looks like a triangle with 1 at the top vertex and running down the two slanting sides.

This array of numbers is known as Pascal’s triangle, after the name of French mathematician Blaise Pascal. It is also known as Meru Prastara by Pingla.

### NCERT Solutions Class 11 Maths Chapter 8 Binomial Theorem

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(1)5 – 5C(1)4 (2x) + 5C(1)(2x)2 – 5C(1)(2x)3 + 5C(1)1 (2x)4 – 5C(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 the binomial theorem, given equation can be expanded as

Solution:

From the binomial theorem, the 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 + 4C(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 C(1.1) + other positive terms

= 1 + 10000 × 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)10000 > 1000