Question

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

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

Answer 1

Answer: (A)

$t=\sum _{r=0}^n\frac{1}{{}^n{C}_r}=\sum _{r=0}^n\frac{n-r}{{}^n{C}_{n-r}}=\sum _{r=0}^n\frac{n-r}{{}^n{C}_{n-r}}\left(\because {}^n{C}_r{=}^n{C}_{n-r}\right)$
$2t_n=\sum _{r=0}^n\frac{r+n-r}{{}^n{C}_r}=\sum _{r=0}^n\frac{n}{{}^n{C}_r}\Rightarrow t_n=\frac{n}{2}\sum _{r=0}^n\frac{1}{{}^n{C}_r}=\frac{n}{2}{S}_n\Rightarrow \frac{t_n}{S_n}=\frac{n}{2}$