 # Important Questions for Class 11 Maths Chapter 8 - Binomial Theorem

Some of the Important questions for class 11 Maths Chapter 8 – Binomial Theorem are provided here. While calculating the higher powers, the multiplication process becomes complex. To avoid the difficulty of repeated multiplication, the binomial theorem was introduced. In class 11 Maths Chapter 8, students can learn about the binomial theorem for positive integral indices. Also, it covers the concept of middle terms. Here, all the important questions from chapter 8 are provided with solutions. These questions are referred from the previous year questions papers and NCERT textbooks. Access, all the chapters important questions for class 11 Maths at BYJU’S.

Class 11 Maths Chapter 8 – Binomial Theorem describe the following important concepts:

• Introduction to binomials
• Binomial theorem for positive integral indices
• For Positive integers “n” and special cases
• General and middle terms

Also, Check:

## Class 11 Chapter 8 – Binomial Theorem Important Questions with Solutions

Practice the given important questions for class 11 Maths Chapter 8 – Binomial Theorem given below.

Question 1:

Expand the expression (2x-3)6 using the binomial theorem.

Solution:

Given Expression: (2x-3)6

By using the binomial theorem, the expression (2x-3)6 can be expanded as follows:

(2x-3)6 = 6C0(2x)66C1(2x)5(3) + 6C2(2x)4(3)26C3(2x)3(3)3 + 6C4(2x)2(3)46C5(2x)(3)5 + 6C6(3)6

(2x-3)6 = 64x6 – 6(32x5 )(3) +15(16x4 )(9) – 20(8x3 )(27) +15(4x2 )(81) – 6(2x)(243) + 729

(2x-3)6 = 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

Thus, the binomial expansion for the given expression (2x-3)6 is 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.

Question 2:

Evaluate (101)4 using the binomial theorem

Solution:

Given: (101)4.

Here, 101 can be written as the sum or the difference of two numbers, such that the binomial theorem can be applied.

Therefore, 101 = 100+1

Hence, (101)4 = (100+1)4

Now, by applying the binomial theorem, we get:

(101)4 = (100+1)4 = 4C0(100)4 +4C1 (100)3(1) + 4C2(100)2(1)2 +4C3(100)(1)3 +4C4(1)4

(101)4 = (100)4+4(100)3+6(100)2+4(100) + (1)4

(101)4 = 100000000+ 4000000+ 60000+ 400+1

(101)4 = 104060401

Hence, the value of (101)4 is 104060401.

Question 3:

Using the binomial theorem, show that 6n–5n always leaves remainder 1 when divided by 25

Solution:

Assume that, for any two numbers, say x and y, we can find numbers q and r such that x = yq + r, then we say that b divides x with q as quotient and r as remainder. Thus, in order to show that 6n – 5n leaves remainder 1 when divided by 25, we should prove that 6n – 5n = 25k + 1, where k is some natural number.

We know that,

(1 + a)n = nC0 + nC1 a + nC2 a2 + … + nCn an

Now for a=5, we get:

(1 + 5)n = nC0 + nC1 5 + nC2 (5)2 + … + nCn 5n

Now the above form can be weitten as:

6n = 1 + 5n + 52 nC2 + 53 nC3+ ….+ 5n

Now, bring 5n to the L.H.S, we get

6n – 5n = 1 + 52 nC2 + 53 nC3+ ….+ 5n

6n – 5n = 1 + 52 (nC2 + 5 nC3+ ….+ 5n-2)

6n – 5n = 1 + 25 (nC2 + 5 nC3+ ….+ 5n-2)

6n – 5n = 1 + 25 k (where k =nC2 + 5 nC3+ ….+ 5n-2)

The above form proves that, when 6n–5n is divided by 25, it leaves the remainder 1.

Hence, the given statement is proved.

Question 4:

Find the value of r, If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal.

Solution:

For the given condition, the coefficients of (r – 5)th and (2r – 1)th terms of the expansion (1 + x)34 are 34Cr-6 and 34C2r-2 respectively.

Since the given terms in the expansion are equal,

34Cr-6 = 34C2r-2

From this, we can write it as either

r-6=2r-2

(or)

r-6=34 -(2r-2) [We know that, if nCr = nCp , then either r = p or r = n – p]

So, we get either r = – 4 or r = 14.

We know that r being a natural number, the value of r = – 4 is not possible.

Hence, the value of r is14.

### Practice Problems for Class 11 Maths Chapter 8

These questions given below are given to make you confident about the topic. You’ll be able to solve all types of class 11 chapter 8 questions if you practice these NCERT and other given questions Properly.

1. Expand (2a – 3b)4 by binomial theorem.
2. Using Binomial theorem, expand (a + 1/b)11.
3. Write the general term in the expansion of (a2 – b )6.
4. Find x, if T11 and T12 in the expansion of (2+ x)50 are equal.
5. The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1:7:42. Find n.