If a_n = sum_{r=0}^n frac{1}{binom{n}{r}},the | vedclass

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(C) Let $S = \sum_{r=0}^n \frac{r}{\binom{n}{r}}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,we can write:$S = \sum_{r=0}^n \frac{n-r}{\binom{

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$\frac{n}{2} a_n$

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(C) Let $S = \sum_{r=0}^n \frac{r}{\binom{n}{r}}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,we can write:$S = \sum_{r=0}^n \frac{n-r}{\binom{n}{n-r}} = \sum_{r=0}^n \frac{n-r}{\binom{n}{r}}$.Adding the two expressions for $S$:$2S = \sum_{r=0}^n \frac{r}{\binom{n}{r}} + \sum_{r=0}^n \frac{n-r}{\binom{n}{r}} = \sum_{r=0}^n \frac{r + n - r}{\binom{n}{r}} = \sum_{r=0}^n \frac{n}{\binom{n}{r}}$.$2S = n \sum_{r=0}^n \frac{1}{\binom{n}{r}} = n a_n$.Therefore,$S = \frac{n}{2} a_n$.

If $a_n = \sum_{r=0}^n \frac{1}{\binom{n}{r}}$,then $\sum_{r=0}^n \frac{r}{\binom{n}{r}} = .....$

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