NCERT solution class 11 chapter 8 Binomial Theorem exercise 8.2 mathematics

EXERCISE 8.2

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Page No 171:

Question 1:

Find the coefficient of x⁵ 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 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 T_r +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 (r + 1)^th term, (T_r+1), in the binomial expansion of (a + 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 b⁷ and in T_r +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 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 (x² – y⁶) is

Question 4:

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

Question 5:

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

Question 6:

Find the 13^th term in the expansion of.

Answer:

It is known that (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 x⁵y⁵.

Question 9:

In the expansion of (1 + a)^{m + n}, prove that coefficients of a^m and aⁿ 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 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 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 T_{k + 1}, we obtain
k = n
Therefore, the coefficient of aⁿ is

Thus, from (1) and (2), it can be observed that the coefficients of a^m and aⁿ 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
4r – 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 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 (r + 1)^th term, (T_r+1), in the binomial expansion of (a + 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 T_{r + 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)^{2n – 1}, we obtain

Comparing the indices of x in xⁿ and T_{k + 1}, we obtain
k = n
Therefore, the coefficient of xⁿ in the expansion of (1 + x)^2n –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 (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by .
Assuming that x² occurs in the (r + 1)^th term of the expansion (1 +x)^m, we obtain

Comparing the indices of x in x² and in T_{r + 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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