Question

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

A. $$\left( {n - 1} \right){a_n}$$

B. $$n{a_n}$$

C. $$\frac{1}{2}n{a_n}$$

D. None of these

Answer : $$\frac{1}{2}n{a_n}$$

Solution :
Take $$n = 2m.$$ Then
$$\eqalign{ & {a_n} = \frac{1}{{^{2m}{C_0}}} + \frac{1}{{^{2m}{C_1}}} + ..... + \frac{1}{{^{2m}{C_{2m}}}} \cr & {a_n} = 2\left( {\frac{1}{{^{2m}{C_0}}} + \frac{1}{{^{2m}{C_1}}} + ..... + \frac{1}{{^{2m}{C_{m - 1}}}}} \right) + \frac{1}{{^{2m}{C_m}}}\,\,\,\,.....\left( 1 \right) \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = \sum\limits_{r = 0}^{2m} {\frac{r}{{^{2m}{C_r}}} = \frac{1}{{^{2m}{C_1}}} + \frac{2}{{^{2m}{C_2}}} + ..... + \frac{{2m}}{{^{2m}{C_{2m}}}}} \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = \left( {\frac{1}{{^{2m}{C_1}}} + \frac{{2m - 1}}{{^{2m}{C_{2m - 1}}}}} \right) + \left( {\frac{2}{{^{2m}{C_2}}} + \frac{{2m - 2}}{{^{2m}{C_{2m - 2}}}}} \right) + ..... + \left( {\frac{{m - 1}}{{^{2m}{C_{m - 1}}}} + \frac{{m + 1}}{{^{2m}{C_{m + 1}}}}} \right) + \frac{m}{{^{2m}{C_m}}} + \frac{{2m}}{{^{2m}{C_{2m}}}} \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = 2m\left\{ {\frac{1}{{^{2m}{C_1}}} + \frac{1}{{^{2m}{C_2}}} + ..... + \frac{1}{{^{2m}{C_{m - 1}}}}} \right\} + \frac{m}{{^{2m}{C_m}}} + 2m \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = 2m\left\{ {\frac{1}{{^{2m}{C_0}}} + \frac{1}{{^{2m}{C_1}}} + \frac{1}{{^{2m}{C_2}}} + ..... + \frac{1}{{^{2m}{C_{m - 1}}}}} \right\} + \frac{m}{{^{2m}{C_m}}} \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = m\left( {{a_n} - \frac{1}{{^{2m}{C_m}}}} \right) + \frac{m}{{^{2m}{C_m}}} = m{a_n} = \frac{n}{2}{a_n}. \cr} $$
Similarly for $$n = 2m + 1.$$

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