## NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Ex 8.2

#### Page No 171:

#### Question 1:

Find the coefficient of

*x*^{5}in (*x*+ 3)^{8}#### 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*^{5}occurs in the (*r*+ 1)^{th}term of the expansion (*x*+ 3)^{8}, we obtain
Comparing the indices of

*x*in*x*^{5}and in*T*_{r}_{ +1}, we obtain*r*= 3

Thus, the coefficient of

*x*^{5}is#### Question 2:

Find the coefficient of

*a*^{5}*b*^{7}in (*a*– 2*b*)^{12}#### 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

*a*^{5}*b*^{7}occurs in the (*r*+ 1)^{th}term of the expansion (*a*– 2*b*)^{12}, we obtain
Comparing the indices of

*a*and*b*in*a*^{5}b^{7 }and in*T*_{r}_{ +1}, we obtain*r*= 7

Thus, the coefficient of

*a*^{5}*b*^{7}is#### Question 3:

Write the general term in the expansion of (

*x*^{2}–*y*)^{6}#### Answer:

It is known that the general term

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

*x*^{2}–*y*^{6}) is#### Question 4:

Write the general term in the expansion of (

*x*^{2}–*yx*)^{12},*x*≠ 0#### Answer:

It is known that the general term

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

*x*^{2}–*yx*)^{12}is#### Question 5:

Find the 4

^{th}term in the expansion of (*x*– 2*y*)^{12 }.#### Answer:

It is known that (

*r*+ 1)^{th}term, (*T*_{r}_{+1}), in the binomial expansion of (*a*+*b*)^{n}is given by .
Thus, the 4

^{th}term in the expansion of (*x*– 2*y*)^{12}is#### Question 6:

Find the 13

^{th}term in the expansion of.#### Answer:

*r*+ 1)

^{th}term, (

*T*

_{r}

_{+1}), in the binomial expansion of (

*a*+

*b*)

^{n}is given by .

Thus, 13

^{th}term in the expansion of is#### Question 7:

Find the middle terms in the expansions of

#### Answer:

It is known that in the expansion of (

*a*+*b*)^{n}, if*n*is odd, then there are two middle terms, namely, term and term.
Therefore, the middle terms in the expansion of are term and term

Thus, the middle terms in the expansion of are .

#### Question 8:

Find the middle terms in the expansions of

#### Answer:

It is known that in the expansion (

*a*+*b*)^{n}, if*n*is even, then the middle term is term.
Therefore, the middle term in the expansion of is term

Thus, the middle term in the expansion of is 61236

*x*^{5}*y*^{5}.#### Question 9:

In the expansion of (1 +

*a*)^{m + n}, prove that coefficients of*a*^{m}and*a*^{n}are equal.#### Answer:

*r*+ 1)

^{th}term, (

*T*

_{r}

_{+1}), in the binomial expansion of (

*a*+

*b*)

^{n}is given by .

Assuming that

*a*^{m}occurs in the (*r*+ 1)^{th}term of the expansion (1 +*a*)^{m}^{ + }^{n}, we obtain
Comparing the indices of

*a*in*a*^{m}and in*T*_{r }_{+ 1}, we obtain*r*=

*m*

Therefore, the coefficient of

*a*^{m}is
Assuming that

*a*^{n}occurs in the (*k*+ 1)^{th}term of the expansion (1 +*a*)^{m}^{+}^{n}, we obtain
Comparing the indices of

*a*in*a*^{n}and in*T*_{k}_{ + 1}, we obtain*k*=

*n*

Therefore, the coefficient of

*a*^{n}is
Thus, from (1) and (2), it can be observed that the coefficients of

*a*^{m}and*a*^{n}in the expansion of (1 +*a*)^{m}^{ + }^{n}are equal.#### Question 10:

The coefficients of the (

*r*– 1)^{th},*r*^{th}and (*r*+ 1)^{th}terms in the expansion of
(

*x*+ 1)^{n}are in the ratio 1:3:5. Find*n*and*r*.#### Answer:

It is known that (

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

*r*– 1)^{th}term in the expansion of (*x*+ 1)^{n}is*r*

^{ th}term in the expansion of (

*x*+ 1)

^{n}is

(

*r*+ 1)^{th}term in the expansion of (*x*+ 1)^{n}is
Therefore, the coefficients of the (

*r*– 1)^{th},*r*^{th}, and (*r*+ 1)^{th}terms in the expansion of (*x*+ 1)^{n}are respectively. Since these coefficients are in the ratio 1:3:5, we obtain
Multiplying (1) by 3 and subtracting it from (2), we obtain

4

*r*– 12 = 0
⇒

*r*= 3
Putting the value of

*r*in (1), we obtain*n*– 12 + 5 = 0

⇒

*n*= 7
Thus,

*n*= 7 and*r*= 3#### Question 11:

Prove that the coefficient of

*x*^{n}in the expansion of (1 +*x*)^{2}^{n}is twice the coefficient of*x*^{n}in the expansion of (1 +*x*)^{2}^{n}^{–1 }.#### Answer:

*r*+ 1)

^{th}term, (

*T*

_{r}

_{+1}), in the binomial expansion of (

*a*+

*b*)

^{n}is given by .

Assuming that

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

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

*n*

Therefore, the coefficient of

*x*^{n}in the expansion of (1 +*x*)^{2}^{n}is
Assuming that

*x*^{n}occurs in the (*k*+1)^{th}term of the expansion (1 +*x*)^{2}^{n }^{– 1}, we obtain
Comparing the indices of

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

*n*

Therefore, the coefficient of

*x*^{n}in the expansion of (1 +*x*)^{2}^{n }^{–1}is
From (1) and (2), it is observed that

Therefore, the coefficient of

*x*^{n}in the expansion of (1 +*x*)^{2}^{n}is twice the coefficient of*x*^{n}in the expansion of (1 +*x*)^{2}^{n}^{–1}.
Hence, proved.

#### Question 12:

Find a positive value of

*m*for which the coefficient of*x*^{2}in the expansion
(1 +

*x*)^{m}is 6.#### Answer:

*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 of the expansion (1 +*x*)^{m}, we obtain
Comparing the indices of

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

Therefore, the coefficient of

*x*^{2}is.
It is given that the coefficient of

*x*^{2}in the expansion (1 +*x*)^{m}is 6.
Thus, the positive value of

*m*, for which the coefficient of*x*^{2}in the expansion
(1 +

*x*)^{m}is 6, is 4._{}

^{}