Question

If $S_n=\sum _{r=0}^n\frac{1}{{}^n{C}_r}$ and $P_n=\sum _{r=0}^n\frac{r}{{}^n{C}_r}$ then $\frac{S_n}{P_n}$ is

• A. $\frac{1}{n}$
• B. $\frac{2}{n}$
• C. $\frac{8}{n}$
• D. $\frac{4}{n}$

Answer 1

The correct option is B $\frac{2}{n}$

$P_n=\sum _{r=0}^n\frac{r}{{}^n{C}_r}\Rightarrow P_n=\sum _{r=0}^n\frac{n-(n-r)}{{}^n{C}_r}\Rightarrow P_n=\sum _{r=0}^n\frac{n}{{}^n{C}_r}-\frac{(n-r)}{{}^n{C}_r}\Rightarrow P_n=n\sum _{r=0}^n\frac{1}{{}^n{C}_r}-\sum _{r=0}^n\frac{(n-r)}{{}^n{C}_{n-r}}\Rightarrow P_n=n{S}_n-\sum _{r=0}^n\frac{r}{{}^n{C}_r}\Rightarrow P_n=n{S}_n-P_n\Rightarrow 2P_n=n{S}_n\Rightarrow \frac{S_n}{P_n}=\frac{2}{n}$