# Binomial Theorem Class 11

## Trigonometry # Binomial Theorem Class 11

In binomial theorem class 11, chapter 8 provides the information regarding the introduction and basic definitions for binomial theorem in a detailed way. To score good marks in binomial theorem class 11 concepts, go through the given problems here. Solve all class 11 Maths Chapter 8 problems in the book by referring to the examples to clarify your binomial theorem concepts.

## Binomial Theorem Class 11 Topics

The topics and sub-topics covered in binomial theorem class 11 are:

• Introduction
• Binomial theorem for positive integral indices
• Binomial theorem for any positive integer n
• Special Cases
• General and Middle Term

## Binomial Theorem Class 11 Notes

The binomial theorem states a formula for the expression of the powers of sums. The most succinct version of this formula is shown immediately below:

$(x+y)^r=\sum_{k=0}^{\infty}\binom{r}{k}x^{r-k}y^k$

From the above representation, we can expand (a + b)n as given below:

(a + b)nnC0 annC1 an-1 b + nC2 an-2 b2 + … + nCn-1 a bn-1nCn bn

This is the binomial theorem formula for any positive integer n.

Some special cases from the binomial theorem can be written as:

• (x + y)nnC0 xnnC1 xn-1 by+ nC2 xn-2 y2 + … + nCn-1 x yn-1nCn xn
• (x – y)nnC0 xn – nC1 xn-1 by + nC2 xn-2 y2 + … + (-1)n nCn xn
• (1 – x)nnC0 – nC1 x + nC2 x2 – …. (-1)n nCn xn

Also, nC0nCn = 1

However, there will be (n + 1) terms in the expansion of (a + b)n.

General and Middle terms

Consider the binomial expansion, (a + b)nnC0 annC1 an-1 b + nC2 an-2 b2 + … + nCn-1 a bn-1nCn bn

Here,

First term = nC0 an

Second term = nC1 an-1 b

Third term = nC2 an-2 b2

Similarly, we can write the (r + 1)th term as:

nCr an-r br

This is the general term of the given expansion.

Thus, (r + 1)Th term, i.e. Tr+1nCr an-r br is called the middle term of the expansion (a + b)n.

### Binomial Theorem Class 11 Examples

Example 1: Expand: [x2 + (3/x)]4, x ≠ 0

Solution:

[x2 + (3/x)]4

Using binomial theorem,

[x2 + (3/x)]44C0 (x2)44C1 (x2)3 (3/x) + 4C2 (x2)2 (3/x)24C3 (x2) (3/x)34C4 (3/x)4

= x8 + 4 x6 (3/x) + 6 x4 (9/x2) + 4 x2 (27/x3) + (81/x4)

= x8 + 12x5 + 54x2 + (108/x) + (81/x4)

Example 2: Compute (98)5

Solution:

Let us write the number 98 as the difference between the two numbers.

98 = 100 – 2

So, (98)5 = (100 – 2)5

Using binomial expansion,

(98)55C0 (100)5 – 5C1 (100)4 (2) + 5C2 (100)3 (2)2 – 5C3 (100)2 (2)35C4 (100) (2)4 – 5C5 (2)5

=  10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32

= 10040008000 – 1000800032

= 9039207968

Example 3: Find the coefficient of x6y3 in the expansion (x + 2y)9.

Solution:

Let x6y3 be the (r + 1)th term of the expansion (x + 2y)9.

So,

Tr+19Cr x9-r (2y)r

x6y39Cr x9-r 2r yr

By comparing the indices of x and y, we get r = 3.

Coefficient of x6y39C3 (2)3

= 84 × 8

= 672

Therefore, the coefficient of x6y3 in the expansion (x + 2y)9 is 672.

Example 4: The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n.

Solution:

Given,

Second term = T2 = 240

Third term = T3 = 720

Fourth term = T4 = 1080

Now,

T2 = T1+1nC1 xn-1 (a)

nC1 xn-1 a = 240….(i)

Similarly,

nC2 xn-2 a2 = 720….(ii)

nC3 xn-3 a3 = 1080….(iii)

Dividing (ii) by (i),

[nC2 xn-2 a2]/ [nC1 xn-1 a] = 720/240

[(n – 1)!/(n – 2)!].(a/x) = 6

(n – 1) (a/x) = 6

a/x = 6/(n – 1)….(iv)

Similarly, by dividing (iii) by (ii),

a/x = 9/[2(n – 2)]….(v)

From (iv) and (v),

6/(n – 1) = 9/[2(n – 2)]

12(n – 2) = 9(n – 1)

12n – 24 = 9n – 9

12n – 9n = 24 – 9

3n = 15

n = 5

Subsituting n = 5 in (i),

5C1 x4 a = 240

ax4 = 240/5

ax4 = 48….(vi)

Substituting n = 5 in (iv),

a/x = 6/(5 – 1)

a/x = 6/4 = 3/2

a = (3x/2)

Putting this oin equ (vi), we get;

(3x/2) x4 = 48

x5 = 32

x5 = 25

⇒ x = 2

Substituting x = 2 in a = (3x/2)

a = 3(2)/2 = 3

Therefore, x = 2, a = 3 and n = 5.

### Practice Problems

1. The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio1: 7 : 42. Find n.
2. What is the middle term in the expansion of [3x – (x3/6)]7?
3. If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r + 1) + 4r2 – 2 = 0.

Stay tuned with BYJU’S – The Learning App to gain knowledge on binomial theorem concepts and also refer to more practice problems to score more marks.