Question

lf $s_n=\sum _{r=0}^n\frac{1}{(C^n{)}_r}$ and $t_n=\sum _{r=0}^n\frac{r}{(C^n{)}_r}$, then $\frac{t_n}{s_n}=$

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

Answer 1

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

$t_n=\sum \frac{r}{n{C}_r}$
$=\sum \frac{n-r}{n{C}_r}$
$=n\sum \frac{1}{n{C}_r}-\sum \frac{r}{n{C}_r}$
$=n{s}_n-t_n$
$t_n=n{s}_n-t_n$
$2t_n=n{s}_n$
$\frac{t_n}{s_n}=\frac{n}{2}$