Question

If $a_n=\sum _{r=n}^n\frac{1}{{}^n{C}_r}$, then $\sum _{r=n}^n\frac{r}{{}^n{C}_r}$ equals?

• A. $(n-1)a_n$
• B. $\frac{{}^n{a}_n}{2}$
• C. $\left(1/2\right)n\cdot a^n$
• D. None of these

Answer 1

Answer: (B)

$\frac{{}^n{a}_n}{2}$

$$\begin{array}{c}let{\sum }_{r=0}^n\frac{r}{{}^n{C}_r}={\sum }_{r=o}^n\frac{n-(n-r)}{{}^n{C}_r}\\ =n.{\sum }_{r=0}^n\frac{1}{{}^n{C}_r}-{\sum }_{r=0}^n\frac{n-r}{{}^n{C}_r}\\ {=}^n{a}_n-{\sum }_{r=0}^n\frac{r}{{}^n{a}_n}\\ 2{\sum }_{r=o}^n\frac{r}{{}^n{a}_n}={}^n{a}_n\\ \therefore {\sum }_{r=0}^n\frac{n}{{}^n{a}_n}=\frac{{}^n{a}_n}{2}\end{array}$$