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

Answer: (A)

Given that, $s_n=\sum _{r=0}^n\frac{1}{{}^n{C}_r}$ $s_n=\sum _{r=0}^n\frac{1}{{}^n{C}_{n-r}}$ $({\because }^n{C}_r{=}^n{C}_{n-r})$ $\Rightarrow$ $n{s}_n=\sum _{r=0}^n\frac{n}{{}^n{C}_{n-r}}$ $\Rightarrow$ $n{s}_n=\sum _{r=0}^n\left[\frac{n-r}{{}^n{C}_{n-r}}+\frac{r}{{}^n{C}_{n-r}}\right]$ $\Rightarrow$ $n{s}_n=\sum _{r=0}^n\frac{n-r}{{}^n{C}_{n-r}}+\sum _{r=0}^n\frac{r}{{}^n{C}_r}$ $\Rightarrow$ $n{s}_n=\left(\frac{n}{{}^n{C}_n}+\frac{n-1}{{}^n{C}_{n-1}}+....+\frac{1}{{}^n{C}_n}\right)+\sum _{r=0}^n\frac{r}{{}^n{C}_r}$ $\Rightarrow$ $n{s}_n=t_n+t_n$ $\Rightarrow$ $n{s}_n=2t_n$ $\Rightarrow$ $\frac{t_n}{s_n}=\frac{n}{2}$