## NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Miscellaneous Exercise

#### Page No 175:

#### Question 1:

Find

*a*,*b*and*n*in the expansion of (*a*+*b*)^{n}if the first three terms of the expansion are 729, 7290 and 30375, respectively.#### Answer:

It is known that (

*r*+ 1)^{th}term, (*T*_{r}_{+1}), in the binomial expansion of (*a*+*b*)^{n}is given by .
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.

Therefore, we obtain

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

From (4) and (5), we obtain

Substituting

*n*= 6 in equation (1), we obtain*a*

^{6}= 729

From (5), we obtain

Thus,

*a*= 3,*b*= 5, and*n*= 6.#### Question 2:

Find

*a*if the coefficients of*x*^{2}and*x*^{3}in the expansion of (3 +*ax*)^{9}are equal.#### Answer:

It is known that (

*r*+ 1)^{th}term, (*T*_{r}_{+1}), in the binomial expansion of (*a*+*b*)^{n}is given by .
Assuming that

*x*^{2}occurs in the (*r*+ 1)^{th}term in the expansion of (3 +*ax*)^{9}, we obtain
Comparing the indices of

*x*in*x*^{2}and in*T*_{r}_{ + 1}, we obtain*r*= 2

Thus, the coefficient of

*x*^{2}is
Assuming that

*x*^{3}occurs in the (*k*+ 1)^{th}term in the expansion of (3 +*ax*)^{9}, we obtain
Comparing the indices of

*x*in*x*^{3}and in*T*_{k}_{+ 1}, we obtain*k*= 3

Thus, the coefficient of

*x*^{3}is
It is given that the coefficients of

*x*^{2}and*x*^{3}are the same.
Thus, the required value of

*a*is.#### Question 3:

Find the coefficient of

*x*^{5}in the product (1 + 2*x*)^{6}(1 –*x*)^{7}using binomial theorem.#### Answer:

Using Binomial Theorem, the expressions, (1 + 2

*x*)^{6}and (1 –*x*)^{7}, can be expanded as
The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve

*x*^{5}, are required.
The terms containing

*x*^{5}are
Thus, the coefficient of

*x*^{5}in the given product is 171.#### Question 4:

If

*a*and*b*are distinct integers, prove that*a*–*b*is a factor of*a*^{n}–*b*^{n}, whenever*n*is a positive integer.
[

**Hint:**write*a*^{n}= (*a*–*b*+*b*)^{n}and expand]#### Answer:

In order to prove that (

*a*–*b*) is a factor of (*a*^{n}–*b*^{n}), it has to be proved that*a*

^{n}–

*b*

^{n}=

*k*(

*a*–

*b*), where

*k*is some natural number

It can be written that,

*a*=*a*–*b*+*b*
This shows that (

*a*–*b*) is a factor of (*a*^{n}–*b*^{n}), where*n*is a positive integer.#### Question 5:

Evaluate.

#### Answer:

Firstly, the expression (

*a*+*b*)^{6}– (*a*–*b*)^{6}is simplified by using Binomial Theorem.
This can be done as

#### Question 6:

Find the value of.

#### Answer:

Firstly, the expression (

*x*+*y*)^{4}+ (*x*–*y*)^{4}is simplified by using Binomial Theorem.
This can be done as

#### Question 7:

Find an approximation of (0.99)

^{5}using the first three terms of its expansion.#### Answer:

0.99 = 1 – 0.01

Thus, the value of (0.99)

^{5}is approximately 0.951.#### Question 8:

Find

*n*, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of#### Answer:

In the expansion, ,

Fifth term from the beginning

Fifth term from the end

Therefore, it is evident that in the expansion of, the fifth term from the beginning is and the fifth term from the end is.

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is. Therefore, from (1) and (2), we obtain

Thus, the value of

*n*is 10.#### Page No 176:

#### Question 9:

Expand using Binomial Theorem.

#### Answer:

Using Binomial Theorem, the given expression can be expanded as

Again by using Binomial Theorem, we obtain

From (1), (2), and (3), we obtain

#### Question 10:

Find the expansion of using binomial theorem.

#### Answer:

Using Binomial Theorem, the given expression can be expanded as

Again by using Binomial Theorem, we obtain

From (1) and (2), we obtain

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^{}