Q. $\sum_{r = 0}^{n}\frac{( - 1)^{r}}{{\ ^{n}C}_{r}}$ equals
A) $\frac{(n + 1)}{(n + 2)}\lbrack 1 + ( - 1)^{n}\rbrack$
B) 0
C) $\frac{2(n + 1)}{(n + 2)}$
D) $\frac{n}{n + 1}\left\lbrack 1 + ( - 1)^{n} \right\rbrack$
Explanation
(a)
$\therefore\ \frac{( - 1)^{r}}{\ ^{n}C_{r}} = \frac{n + 1}{n + 2}\left( \frac{( - 1)^{r}}{\ ^{n + 1}C_{r + 1}} - \frac{( - 1)^{r - 1}}{\ ^{n + 1}C_{r}} \right)$
$\Rightarrow \ \ \sum_{r = 0}^{n}{a_{r} =}\frac{n + 1}{n + 2}\left( 1 + ( - 1)^{n} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.