If $a_n=\sum _{r=0}^n \frac{1}{{ }^n{C}_r}$, then the value of $\sum _{r=0}^n \frac{r}{{ }^n{C}_r}$

Let $P=\sum _{r=0}^n \frac{r}{{ }^n{C}_r}$                     ….(i)

Replacing $r$ by $(n - r)$ in Eq. (i), we get 

$P=\sum _{r=0}^n \frac{(n-r)}{{ }^n{C}_{n-r}}=\sum _{r=0}^n \frac{(n-r)}{{ }^n{C}_r}\left[{\because }^n{C}_r{=}^n{C}_{n-r}\right]\dots \text{ (ii) }$

On adding Eqs. (i) and (ii), we get

$2P=\sum _{r=0}^n \frac{n}{{ }^n{C}_r}=n\sum _{r=0}^n \frac{1}{{ }^n{C}_r}=n{a}_n$ [given]

$\therefore P=\frac{n}{2}{a}_n$

Hence, $\sum _{r=0}^n \frac{r}{{ }^n{C}_r}=\frac{n}{2}{a}_n$