If $S_n =\displaystyle\sum_{r=0}^{n} \frac{1}{^nC_r}$ and $t_n= \displaystyle\sum_{r=0}^{n} \frac{r}{^nC_r}$, then $\frac{t_n}{S_n}$ is equal to

Question

( A ) $\frac{1}{2}n-1$
( B ) $\frac{1}{2}n$
( C ) $\frac{2n-1}{2}$
( D ) $n-1$