Question

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

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

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

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

$=n\sum \frac{1}{{}^n{C}_r}-\sum _{r=0}^n\frac{r}{{}^n{C}_r}$ replacing $n-r$ by $r$

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

$\therefore 2t_n=n{S}_n$

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