Q. $If\ {(1 + x)}^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ... + C_{n}x^{n}.$Then , $C_{0}C_{1} + C_{1}C_{2} + ... + C_{n - 1}C_{n}\$is equal to
A) $\frac{(2n)!}{(n - 1)!(n + 1)!}$
B) $\frac{(2n - 1)!}{(n - 1)!(n + 1)!}$
C) $\frac{2n!}{(n + 2)!(n + 1)!}$
D) None of these
Explanation
(a)
Given, $(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + \ldots + C_{n}x^{n}$
Also, $(x + 1)^{n} = C_{n} + C_{n - 1}x + C_{n - 2}x^{2} + \ldots + C_{0}x^{n}$
On multiplying both equations and comparing coefficient of $x^{n - 1}$ on both sides, we get
$C_{0}C_{2} + C_{1}C_{2} + C_{2}C_{3} + \ldots + C_{n - 1}C_{n} = \ ^{2n}C_{n - 1}$
$= \frac{(2n)!}{(n - 1)!(n + 1)!}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.