Chapter 8 Binomial Theorem Ex 8.2 Solutions
Question - 1 : - Find the coefficient of x5 in (x + 3)8
Answer - 1 : -
It is known that (r +1)th term, (Tr+1), in the binomialexpansion of (a + b)n is givenby 
.
Assuming that x5 occursin the (r + 1)th term of the expansion (x +3)8, we obtain 
Comparing the indices of x in x5 andin Tr +1, we obtain
r = 3
Thus, the coefficient of x5 is 
Question - 2 : - Find the coefficient of a5b7 in(a – 2b)12
Answer - 2 : -
It is known that (r +1)th term, (Tr+1), in the binomialexpansion of (a + b)n is givenby .
Assuming that a5b7 occursin the (r + 1)th term of the expansion (a –2b)12, we obtain
Comparing the indices of a and b in a5 b7 andin Tr +1, we obtain
r = 7
Thus, the coefficient of a5b7 is 
Question - 3 : - Write the general term in theexpansion of (x2 – y)6
Answer - 3 : -
It is known that the generalterm Tr+1 {which is the (r +1)th term} in the binomial expansion of (a + b)n isgiven by .
Thus, the general term in theexpansion of (x2 – y6) is
Question - 4 : - Write the general term in theexpansion of (x2 – yx)12, x ≠ 0
Answer - 4 : -
It is known that the generalterm Tr+1 {which is the (r +1)th term} in the binomial expansion of (a + b)n isgiven by .
Thus, the general term in theexpansion of(x2 – yx)12 is
Question - 5 : - Find the 4th termin the expansion of (x – 2y)12 .
Answer - 5 : -
It is known that (r +1)th term, (Tr+1), in the binomialexpansion of (a + b)n is givenby .
Thus, the 4th termin the expansion of (x – 2y)12 is
Question - 6 : - Find the 13th term in the expansion of 
.
Answer - 6 : -
It is known that (r +1)th term, (Tr+1), in the binomialexpansion of (a + b)n is givenby .
Thus, 13th termin the expansion of 
is
Question - 7 : - Find the middle terms in the expansions of 
Answer - 7 : -
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.Thus, the middle terms in the expansion of
are 
.
Question - 8 : - Find the middle terms in the expansions of 
Answer - 8 : -
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 
termThus, the middle term in theexpansion of is 61236 x5y5.
Question - 9 : - In the expansion of (1 + a)m+ n, prove that coefficients of am and an areequal.
Answer - 9 : -
It is known that (r +1)th term, (Tr+1), in the binomialexpansion of (a + b)n is givenby .
Assuming that am occursin the (r + 1)th term of the expansion (1 + a)m + n,we obtain
Comparing the indices of a in am andin Tr + 1, we obtain
r = m
Therefore, the coefficientof am is
Assuming that an occursin the (k + 1)th term of the expansion (1 + a)m+n,we obtain
Comparing the indices of a in an andin Tk + 1, we obtain
k = n
Therefore, the coefficientof an is
Thus, from (1) and (2), it can beobserved that the coefficients of am and an inthe expansion of (1 + a)m + n areequal.
Question - 10 : - The coefficients of the (r –1)th, rth and (r + 1)th termsin the expansion of
(x + 1)n arein the ratio 1:3:5. Find n and r.
Answer - 10 : -
It is known that (k +1)th term, (Tk+1), in the binomialexpansion of (a + b)n is givenby 
.
Therefore, (r – 1)th termin the expansion of (x + 1)n is
r th termin the expansion of (x + 1)n is 
(r + 1)th termin the expansion of (x + 1)n is 
Therefore, the coefficients ofthe (r – 1)th, rth, and (r +1)th terms in the expansion of (x + 1)n are 
respectively. Since thesecoefficients are in the ratio 1:3:5, we obtain
Multiplying (1) by 3 andsubtracting 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 = 7and r = 3