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:

\(\begin{array}{l}(x+y)^r=\sum_{k=0}^{\infty}\binom{r}{k}x^{r-k}y^k\end{array} \)

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.

Learn more about the general and middle terms of the binomial expansion here.

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.

Related Links
Complex Number Class 11 Limits and Derivatives Class 11
Linear Inequalities Class 11 Permutation and combination class 11
Test your knowledge on Binomial Theorem Class 11

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