If ${a_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}},}$ then ${a_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}},}$ equals 

$\begin{aligned} \text { Let, } b & =\sum_{r=0}^{n} \frac{r}{n c_{r}}=(1) \\ & =\sum_{r=0}^{n} \frac{n-\pi}{n_{c_{n-\pi}}} \quad \begin{array}{r} \text {[ replace r } \\ \text { by } n-r] \end{array} \\ b & =\sum_{r=0}^{n} \frac{n-\pi}{n_{c_{r}}}-(2) \end{aligned}$

Adding (1) $\&(2) \Rightarrow$

$\begin{aligned} 2 b & =\sum_{r=0}^{n} \frac{r}{n_{c R}}+\sum_{r=0}^{n} \frac{n-r}{n_{c_{R}}} \\ & =n \sum_{R=0}^{n} \frac{1}{n_{c R}} \\ b & =\frac{n a_{n}}{2} \text { [Ans.] } \end{aligned}$