Q. If in the expansion of ${(1 + x)}^{n}$, the coefficient of $r$th and $(r + 2)$th term be equal, then $r$ is equal to
A) $2n$
B) $\frac{2n + 1}{2}$
C) $\frac{n}{2}$
D) $\frac{2n - 1}{2}$
Explanation
(c)
Given that, $\ ^{n}C_{r - 1} = \ ^{n}C_{r + 1}$
$\Rightarrow \frac{n!}{(n - r + 1)(n - r)(n - r - 1)!(r - 1)!}$
$= \frac{n!}{(n - r - 1)!(r + 1)(r)(r - 1)!}$
$\Rightarrow r^{2} + r = n^{2} - nr + n - nr + r^{2} - r$
$\Rightarrow n^{2} - 2nr - 2r + n = 0$
$\Rightarrow (n - 2r)(n + 1) = 0\ \Rightarrow \ r = \frac{n}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.