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NCERT solution class 11 chapter 8 Binomial Theorem exercise 8.2 mathematics

EXERCISE 8.2


Page No 171:

Question 1:

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

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain

Comparing the indices of x in x5 and in Tr +1, we obtain

r = 3

Thus, the coefficient of x5 is

Question 2:

Find the coefficient of a5b7 in (a – 2b)12

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain

Comparing the indices of a and b in a5 band in Tr +1, we obtain

r = 7

Thus, the coefficient of a5b7 is 

Question 3:

Write the general term in the expansion of (x2 – y)6

Answer:

It is known that the general term Tr+1 {which is the (+ 1)th term} in the binomial expansion of (b)n is given by .

Thus, the general term in the expansion of (x2 – y6) is

Question 4:

Write the general term in the expansion of (x2 – yx)12x ≠ 0

Answer:

It is known that the general term Tr+1 {which is the (+ 1)th term} in the binomial expansion of (b)n is given by .

Thus, the general term in the expansion of(x2 – yx)12 is

Question 5:

Find the 4th term in the expansion of (x – 2y)12 .

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Thus, the 4th term in the expansion of (x – 2y)12 is

Question 6:

Find the 13th term in the expansion of.

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Thus, 13th 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 x5y5.

Question 9:

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

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m + n, we obtain

Comparing the indices of a in am and in T+ 1, we obtain

r = m

Therefore, the coefficient of am is

Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, we obtain

Comparing the indices of a in an and in Tk + 1, we obtain

k = n

Therefore, the coefficient of an is

Thus, from (1) and (2), it can be observed that the coefficients of am and an in the expansion of (1 + a)m + n are equal.

Question 10:

The coefficients of the (r – 1)thrth 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 (+ 1)th term, (Tk+1), in the binomial expansion of (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)thrth, 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– 12 = 0

⇒ r = 3

Putting the value of r in (1), we obtain

n – 12 + 5 = 0

⇒ n = 7

Thus, = 7 and r = 3

Question 11:

Prove 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 .

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain

Comparing the indices of x in xn and in Tr + 1, we obtain

r = n

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

Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2– 1, we obtain

Comparing the indices of x in xn and Tk + 1, we obtain

k = n

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

From (1) and (2), it is observed that

Therefore, 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.

Hence, proved.

Question 12:

Find a positive value of m for which the coefficient of x2 in the expansion

(1 + x)m is 6.

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .

Assuming that x2 occurs in the (+ 1)th term of the expansion (1 +x)m, we obtain

Comparing the indices of x in x2 and in Tr + 1, we obtain

r = 2

Therefore, the coefficient of x2 is.

It is given that the coefficient of x2 in the expansion (1 + x)m is 6.

Thus, the positive value of m, for which the coefficient of x2 in the expansion

(1 + x)m is 6, is 4.


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