# NCERT Solutions for Class 11 Maths Chapter 8

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (Dwipad Pramey) Exercise 8.1, Exercise 8.2 or Miscellaneous Exercise to view online or download in PDF format free for session 2020-21.

UP Board Students can download UP Board solutions for class 11 Maths chapter 8 here in Hindi Medium. Join the discussion Forum to ask your doubts and share your knowledge with your friends. NCERT Solutions Offline Apps are based on new NCERT Books 2020-21 following the latest CBSE Syllabus 2020-2021.

## NCERT Solutions for Class 11 Maths Chapter 8

 Class: 11 Subject: Maths Chapter 8: Binomial Theorem

### 11th Maths Chapter 8 Solutions

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem all exercises including miscellaneous exercise are given below to download in PDF form updated for new session 2020-2021. NCERT Solutions 2020-21 are based on latest CBSE Curriculum 2020-2021 for CBSE Board, UP Board, Uttarakhand, Bihar Board, etc. who are following NCERT Books 2020-21 for their study.

• ### 11th Maths Chapter 8 Solutions in PDF

#### Class 11 Maths Chapter 8 Solution in Hindi Videos

Class 11 Maths Chapter 8 Exercise 8.1 Solution
Class 11 Maths Chapter 8 Exercise 8.2 Solution

#### How to find Middle Term?

The Middle Term: In the expansion of (a + b)^n, the total number of terms are (n + 1). The middle term in the expansion of (a + b)^n depend on n.
1. When n is even: Let n = 2m, where m is positive integer. The total number of terms will be 2m + 1. Hence, the middle term of the expansion (a + b)^n will be 1/2[(2m + 1) + 1], i. e. when n is even then (m + 1)th term or (n/2 + 1)th will be the middle term.
2. When n is odd: Let n = 2m + 1, where m is a positive integer. In the expansion of (a + b)^n the total number of terms will be (m + 2). The middle term in the expansion of (a + b)^n will be (m + 1)th and (m + 2)th term or (n + 1)/2th and ((n + 3)/2th term.

##### Important Terms on Binomial Theorem

1. Binomial Expression: Any expression containing two terms combined by + or – is called Binomial expression. For example: x + 3, 2x + y, x – 4y, 4 – 100x, y – 4, etc.
2. In the expansion of (a + b)^n, the coefficient of first term = coefficient of last term, coefficient of second term = coefficient of second term from last. Thus we get that in the expansion of (a + b)^n, the terms from first term and from the last term at equal distance have the same coefficients.
3. The General Term: The term (r + 1) is called the general term of the expansion (a + b)^n because we can get different terms from this term by giving different values to r. This general term is denoted by Tr+1.

### Important Questions on 11th Maths Chapter 8

Using Binomial theorem,indicate which is larger (1.1)^10000 or 1000.
By splitting 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
= C(10000, 0) + C(10000, 1) (10000)^1 (0.1)^1 + Other positive terms.
= 1 + 10000×0.1 + Other positive terms.
= 1001 + Other positive terms.
> 1000
Hence,(1.1)^10000 > 1000.
Find an approximation of (0.99)^5 using the first three terms of its expansion.
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 0.99 = 1 – 0.01
Therefore,(0.99)^5=(1-0.01)^5
= C(5, 0) (-0.01)^0 + C(5, 1) (-0.01)^1 + C(5, 2) (-0.01)^2 + ⋯
= 1 – 5(0.01) + 10(0.0001) + ⋯
≈ 1 – 0.05 + 0.001
= 0.951
Find the expansion of (3x^2 – 2ax + 3a^2 )^3 using binomial theorem.
Here,(3x^2-2ax+3a^2 )^3=[(3x^2-2ax)-3a^2 ]^3
Using Binomial theorem,the expression [(3x^2-2ax)-3a^2 ]^3 can be expressed as
C(3, 0) (3x^2-2ax)^3 + C(3, 1) (3x^2-2ax)^2 (3a^2) + C(3, 2)(3x^2-2ax)^1 (3a^2)^2 + C(3, 3)(3a^2)^3
=(3x^2-2ax)^3+3(9x^4+4a^2 x^2-12ax^3 )(3a^2 )+3(3x^2-2ax)(9a^4 )+(27a^6 )^
=(3x^2-2ax)^3+(81a^2 x^4+36a^4 x^2-108a^3 x^3 )+(81a^4 x^2-54a^5 x)+〖27a^6〗^ …(1)
Again applying Binomial theorem, we have
(3x^2-2ax)^3
= C(3, 0)(3x^2 )^3 + C(3, 1)(3x^2 )^2 (-2ax)^1 + C(3, 2)(3x^2)(-2ax)^2 + C(3, 3)(-2ax)^3
= 27x^6-3×18ax^5+3×12a^2 x^4-8a^3 x^3
= 27x^6-54ax^5+36a^2 x^4-8a^3 x^3
Now putting the value of (3x^2-2ax)^3 in (1),we have
(3x^2-2ax+3a^2 )^3
= [27x^6-54ax^5+36a^2 x^4-8a^3 x^3 ]+(81a^2 x^4+36a^4 x^2-108a^3 x^3 )+(81a^4 x^2-54a^5 x)+〖27a^6〗
= 27x^6-54ax^5+117a^2 x^4-116a^3 x^3+117a^4 x^2-54a^5 x+27a^6             