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

Answer 1

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

Given, $S_n=\sum _{r=0}^n\frac{1}{{}^n{C}_r}$ and $t_n=\sum _{r=0}^n\frac{r}{{}^n{C}_r}$

$\Rightarrow n{S}_n=\sum _{r=0}^n\frac{n}{{}^n{C}_{n-r}}$

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

$\Rightarrow n{S}_n=\sum _{r=0}^n\left[\frac{n-r}{{}^n{C}_{n-r}}\right]+\sum _{r=0}^n\left[\frac{r}{{}^n{C}_r}\right]$

$\Rightarrow n{S}_n=[\frac{n}{{}^n{C}_n}+\frac{n-1}{{}^n{C}_{n-1}}+\cdots +\frac{0}{{}^n{C}_n}]+\sum _{r=0}^n\frac{r}{{}^n{C}_r}$

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

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