Q. If $C_{0},\ C_{1},\ C_{2},\ \ldots\ldots C_{n}$ denote the binomial coefficient in the expansion of ${(1 + x)}^{n}$, then $C_{0}\frac{C_{1}}{2} + \frac{C_{2}}{3} + ... + \frac{C_{n}}{n + 1}\ is\ equal\ to$
A) $\frac{2^{n + 1} - 1}{n + 1}$
B) $\frac{2^{n} - 1}{n}$
C) $\frac{2^{n - 1} - 1}{n - 1}$
D) $\frac{2^{n + 1} - 1}{n + 2}$
Explanation
(a)
We know that ,
$(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + \ldots + C_{n}x^{n}$
On integrating both sides, from 0 to 1, we get
$\left\lbrack \frac{(1 + x)^{n + 1}}{n + 1} \right\rbrack_{0}^{1} = \left\lbrack {C_{0}x + \frac{C_{1}x^{2}}{2} + \frac{C_{2}x^{3}}{3} + \ldots + \frac{C_{n}x^{n + 1}}{n + 1}}^{\ } \right\rbrack_{0}^{1}$
$\Longrightarrow \frac{2^{n + 1} - 1}{n + 1} = C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + ... + \frac{C_{n}}{n + 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.