Question

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. (n/2)a_n
• D. None

Answer 1

The correct option is C

(n/2)a_n

Let n = 2k+1, then

$a_n=\frac{1}{C_0}+\frac{1}{C_1}+....+\frac{1}{C_k}+\frac{1}{C_{k+1}}+\frac{1}{C_{k+2}}+.....+\frac{1}{C_{2k+1}}$

= $\sum _{r=0}^k\frac{1}{C_r}+\frac{1}{C_{2k+1-r}}=2\sum _{r=0}^k\frac{1}{C_r}$ .

$\sum _{r=0}^{2k+1}\frac{r}{C_r}=\frac{0}{C_0}+\frac{1}{C_1}+\frac{2}{C_2}+.....$

.... + $\frac{k}{C_k}+\frac{k+1}{C_{k-1}}+...+\frac{2k+1}{C_{2k+1}}$

= $\sum _{r=0}^k\frac{2k+1}{C_r}=(2k+1)\sum _0^k\frac{1}{c_r}=(2n+1)\frac{a_n}{2}$

$\therefore$ $\sum _{r=0}^n\frac{r}{C_r}=\frac{n}{2}{a}_n.$