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. $n{a}_n$
• C. $\frac{1}{2}n{a}_n$
• D. $\frac{(n-1)}{2}{a}_n$

Answer 1

The correct option is C $\frac{1}{2}n{a}_n$

We have $a_n=\sum _{r=0}^n\frac{1}{{}^n{C}_r}=\sum _{r=0}^{\frac{n}{2+1}}\frac{1}{{}^n{C}_r}+\frac{1}{{}^n{C}_{n-r}}$
$\Rightarrow a_n=\sum _{r=0}^{\frac{n+1}{2}}\frac{2}{{}^n{C}_r}=2\sum _{r=0}^{\frac{n+1}{2}}\frac{1}{{}^n{C}_r}$
$\therefore \sum _{r=0}^n\frac{1}{{}^n{C}_r}=\sum _{r=0}^{\frac{n+1}{2}}\frac{r+(n-r)}{{}^n{C}_r}$
$\Rightarrow n\sum _{r=0}^n\frac{1}{{}^n{C}_r}=n\left(\frac{a_n}{2}\right)=\frac{n}{2}{a}_n$