Question
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
A.
$$n = 2r$$
B.
$$n = 2r + 1$$
C.
$$n = 3r$$
D.
none of these
Answer :
$$n = 2r$$
Solution :
Given that $$r$$ and $$n$$ are +ve integers such that $$r > 1, n > 2$$
Also in the expansion of $${\left( {1 + x} \right)^{2n}}$$
co - eff. of $${\left( {3r} \right)^{th}}$$ term = co - eff. of $${\left( {r + 2} \right)^{th}}$$ term
$$\eqalign{
& \Rightarrow \,{\,^{2n}}{C_{3r - 1}} = {\,^{2n}}{C_{r + 1}} \cr
& \Rightarrow \,\,3r - 1 = r + 1\,\,{\text{or }}3r - 1 + r + 1 = 2n \cr
& \Rightarrow \,\,r = 1\,\,{\text{or }}2r = n \cr
& {\text{But }}\,r > 1 \cr
& \therefore \,\,n = 2r \cr} $$