Question

If $an=\sum _{r=0}^n\frac{1}{{}^n{C}_r}$ 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. $Noneofthese$

Answer 1

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

$\begin{array}{cc}\text{Solution- Given that}& \\ a_n& =\sum _{r=0}^n\frac{1}{n_n}\\ S& =\sum _{r=0}^n\frac{n}{{}^n{C}_{\mu }}\\ & =\sum _{n=0}^n\frac{n-(n-\mu )}{{}^n{C}_n}=\sum _{n=0}^n\frac{n}{{}^n{C}_{\mu }}-\sum _{n=0}^n\frac{n-n}{{}^n{C}_{\mu }}\end{array}$

As we know that

${}^n{C}_n={}^n{C}_{n-\mu }$

$\Rightarrow S=n{\sum }_{r=0}^n\frac{1}{n_n}-{\sum }_{n=0}^n\frac{n-n}{n_{n-n}}$

$\begin{array}{c}S=n\cdot \left(an\right)-S\\ 2S=n\left(an\right)\\ S=\frac{1}{2}n{a}_n\\ \text{Hence,}\left(C\right)\text{is the correct option.}\end{array}$