Class 11 NCERT Solutions – Chapter 8 Binomial Theorem – Exercise 8.2

Question 1. Find the coefficient of x⁵ in (x+3)⁸

Solution: 

The (r+1)^th term of (x+3)⁸ is given by T_r+1 = ⁸C_r(x)^8-r(3)^r (eq1).

Therefore for x⁵ we need to get 8-r =5 (Because we need to find x⁵. Therefore, power ox must be equal to 5)

So we get r=3.

Now, put r=3 in eq1. We get,

Coefficient of x⁵ = ⁸C₃(x)⁵(3)³ = 8!*3³/(4!*4!) = 1512

Coefficient of x⁵ is 1512.

Question 2. Find the coefficient of a⁵b⁷ in (a-2b)¹².

Solution: 

The (r+1)^th term of (a-2b)¹² is given by T_r+1 = ¹²C_r(a)^12-r(-2b)^r × (eq1)

In the question it is given that exponent of b is 7. Therefore, r should be equal to 7.

By putting r=7 in eq1, we get

Coefficient of a⁵b⁷ = ¹²C₇(-2)⁷ = -101376

Question 3. Write the general term in the expansion of (x²−y)⁶.

Solution: 

General term of the equation (a+b)ⁿ is given as T_r+1 = ⁿC_r(a)^n-r(b)^r.

In this question a= x² and b=-y. After putting the value of a, b and n, we get the general term as

T_r+1 = ⁶C_r(x²)^(12-r)(-y)⁶ = (-1)^{r 6}C_r(x)^(12-2r)(y)^r.

Question 4. Write the general term in the expansion of (x²-yx)¹².

Solution:

General term of the equation (a+b)ⁿ is given as T_r+1 = ⁿC_r(a)^n-r(b)^r.

In this question a= x² and b=-yx. After putting the value of a, b and n, we get the general term as

T_r+1 = ¹²C_r(x²)^(12-r)(-yx)^r = (-1)^{r 12}C_r(x)^(24-2r)(y)^r(x)^r = (-1)^{r 12}C_r(x)^(24-r)(y)^r 

Question 5. Find the 4^th term in the expansion of (x-2y)¹²

Solution:

General term in the expansion of (a+b)ⁿ is written as T_r+1 = ⁿC_r(a)^n-r(b)^r

In the question we are given that a = x, b = -2y and n=12.

To get the 4^th term, we need to put r = 3 (Because r+1=4, therefore r=3).

Therefore, T₄ = ¹²C₃(x)^12-3(-2y)³ = −1760x⁹y³

Question 6. Find the 13^th term in the expansion of (9x-1/3√x)¹⁸

Solution: 

General term in the expansion of (a+b)ⁿ is written as T_r+1 = ⁿC_r(a)^n-r(b)^r

In this question a = 9x, b= -1/3√x and n=18.

To get the 13^th term, we need to put r=12 (Because r+1=13, therefore r=12).

Therefore, T₁₃ = ¹⁸C₁₂(9x)^18-12(-1/3√x)¹² = 18564

Question 7. Find the middle terms in the expansion of (3−x³/6)⁷.

Solution: 

In the expansion of (a+b)ⁿ, if n is odd, then there are two middle terms, namely, ((n+1)/2)^th and ((n+1)/2+1)^th term.

Therefore, middle terms in the expansion of (3−x³/6)⁷are 4^th term and 5^th term.

T₄ = T₃₊₁ = ⁷C₃(3)^7-3(−x³/6)³ = (-1)³7!3⁴x⁹/4!.3!6³=-105x⁹/8

T₅ = T₄₊₁ = ⁷C₄(3)^7-4(−x³/6)⁴= (-1)⁴7!3³x¹²/4!3!6⁴ = 35x¹²/48

Thus, the middle terms are -105x⁹/8 and 35x¹²/48.

Question 8. Find the middle terms in the expansion of (x/3+9y)¹⁰

Solution: 

In the expansion of (a+b)ⁿ, if n is even, then the middle term is (n/2+1)^th term.

Therefore, the middle term is 6^th term.

T₆ = T₅₊₁ = ¹⁰C₅(x/3)⁵(9y)⁵= (10!x⁵9⁵y⁵)/(5!5!3⁵) = 61236x⁵y⁵

Thus, the middle term is 61236x⁵y⁵

Question 9. In the expansion of (1+a)^m+n, prove that coefficients of a^m and aⁿ are equal.

Solution: 

Let us assume that a^m occurs in the (r+1)^th term, we obtain

T_r+1 = ^m+nC_r(1)^m+n-r(a)^r = ^m+nC_ra^r

Comparing the indices of a in a^m and in T_r+1, we obtain r=m

Therefore, the coefficient of a^m is ^m+nC_m = (m+n)!/m!n! …(1)

Let us assume that aⁿ occurs in the (k+1)^th, we obtain

T_k+1 = ^m+nC_k(1)^m+n-k(a)^k = ^m+nC_ka^k

Comparing the indices of a in aⁿ and T_k+1, we obtain k-n

Therefore, the coefficient of aⁿ is ^m+nC_n = (m+n)!/m!n! ….(2)

Thus, from (1) and (2), it can be obtained that the coefficients of a^m and aⁿ 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 of 1:3:5. Find n and r.

Solution: 

(r-1)^th term in the expansion of (x+1)ⁿ is T_r-1 = ⁿC_r-2(x)^n-(r-2)(1)^r-2 = ⁿC_r-2(x)^n-r+2

r^th term in the expansion of (x+1)ⁿ is T_r = ⁿC_r-1(x)^n-(r-1)(1)^r-1 = ⁿC_r-1(x)^n-r+1

(r+1)^th term in the expansion of (x+1)ⁿ is T_r+1= ⁿC_r(x)^n-r(1)^r = ⁿC_r(x)^n-r

Therefore the coefficients of (r-1)^th , r^th and (r+1)^th in the expansion of (x+1)ⁿ are ⁿC_r-2, ⁿC_r-1 and ⁿC_r respectively.

Since these coefficients are in the ratio of 1:3:5, we obtain

ⁿC_r-2/ⁿC_r-1 = 1/3 and ⁿC_r-1/ⁿC_r = 3/5

Solving these two equations we get n-4r+5=0 and 3n-8r+3=0.

After solving these two equations, we get n=7 and r=3

Thus n=7 and r=3.

Question 11. Prove that the coefficient of x^n in the expansion of (1+x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1+x)^2n-1.

Solution: 

In the expansion of (1+x)²ⁿ, T_n+1 = ²ⁿC_n(1)^2n-n(x)ⁿ = ²ⁿC_nxⁿ

Therefore, coefficient of xⁿ in the expansion of (1+x)²ⁿ is ²ⁿC_n

²ⁿC_n = (2n)!/(n!)² ….(1)

Similarly, coefficient of xⁿ in the expansion of (1+x)^2n-1 is ^2n-1C_n

^2n-1C_n = (2n)!/2.(n!)² …(2)

From (1) and (2), 2.^2n-1C_n = ²ⁿC_n

Hence, it is proved that the coefficient of xⁿ in the expansion of (1+x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1+x)^2n-1

Question 12. Find a positive value of m for which the coefficient of x² in the expansion of (1+x)^m is 6.

Solution: 

General term T_r+1 = ^mC_r(1)^m-r(x)^r = ^mC_r(x)^r

Comparing the indices of x in x² and T_r+1, we get r=2

Therefore, ^mC₂ = 6

= 6

m!/(m-2)!=12

m(m-1) = 12=> m²-m-12 = 0

(m-4)(m+3) = 0

m cannot be negative. Therefore, m=4