If S_n=_r=0^n1nr and t_n=_r=0^nrnr, then t_nS_n is equal to:

Question

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

Accepted Answer

Since $\binom{n}{r}=\binom{n}{n-r}$, reversing the summation index in $t_n$:
$$t_n=\sum_{r=0}^n \frac{r}{\binom nr}=\sum_{r=0}^n\frac{n-r}{\binom{n}{n-r}}=\sum_{r=0}^n\frac{n-r}{\binom nr}=nS_n-t_n.$$

$$2t_n=nS_n \Rightarrow \frac{t_n}{S_n}=\frac n2.$$

Solutions · 1