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

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

solution:-

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

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

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

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

(replacing n -r by r)

$t_n{=}^n{S}_n-tn$

$\therefore \frac{tn}{5n}=\frac{n}{2}$

option A is correct.