Q. If $a_{n} = \sum_{r = 0}^{n}\frac{1}{\ ^{n}C_{r}}$, then $\sum_{r = 0}^{n}\frac{r}{\ ^{n}C_{r}}$ equals
A) $(n - 1)a_{n}$
B) $na_{n}$
C) $\frac{1}{2}na_{n}$
D) None of these
Explanation
(c)
Let $b = \sum_{r = 0}^{n}\frac{r}{\ ^{n}C_{r}} = \sum_{r = 0}^{n}\frac{n - (n - r)}{\ ^{n}C_{r}}$
$= n\sum_{r = 0}^{n}\frac{1}{\ ^{n}C_{r}} - \sum_{r = 0}^{n}\frac{n - r}{\ ^{n}C_{r}}$
$= na_{n} - \sum_{r = 0}^{n}\frac{n - r}{\ ^{n}C_{n - r}}\ \ (\because\ ^{n}C_{r} = \ ^{n}C_{n - 1}\ )$
$= na_{n} - b$
$\Rightarrow \ 2b = na_{n}\ \Rightarrow \ b = \frac{n}{2}a_{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.