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 $\frac{t_n}{s_n}=$

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

Answer 1

The correct option is A $\frac{1}{2}n$

$S_n=\sum _{r=0}^n\frac{1}{{}^n{C}_r}=\sum _{r=0}^n\frac{1}{{}^n{C}_{n-r}}\left({\because }^n{C}_r{=}^n{C}_{n-r}\right)$

$t_n=\sum _{r=0}^n\frac{r}{{}^n{C}_{n-r}}$

$\therefore n{S}_n=\sum _{r=0}^n[\frac{n-r}{{}^n{C}_{n-r}}+\frac{r}{{}^n{C}_{n-r}}]$

$=\sum _{r=0}^n\frac{n-r}{{}^n{C}_{n-r}}+\sum _{r=0}^n\frac{r}{{}^n{C}_r}$

$=(\frac{n}{{}^n{C}_n}+\frac{n-1}{{}^n{C}_{n-1}}+...+\frac{0}{{}^n{C}_n})+\sum _{r=0}^n\frac{r}{{}^n{C}_r}$

$\Rightarrow n{S}_n=t_n+t_n=2t_n$

$\Rightarrow \frac{t_n}{S_n}=\frac{n}{2}$