If a_n = sum_{r=0}^n frac{1}{^nC_r},then sum_ | vedclass

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(C) Given $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$.Let $b_n = \sum_{r=0}^n \frac{r}{^nC_r}$.Using the property $^nC_r = ^nC_{n-r}$,we can write:$b_n = \fr

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$\frac{1}{2}na_n$

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(C) Given $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$.Let $b_n = \sum_{r=0}^n \frac{r}{^nC_r}$.Using the property $^nC_r = ^nC_{n-r}$,we can write:$b_n = \frac{0}{^nC_0} + \frac{1}{^nC_1} + \frac{2}{^nC_2} + \dots + \frac{n}{^nC_n}$.Also,$b_n = \frac{n}{^nC_n} + \frac{n-1}{^nC_{n-1}} + \dots + \frac{0}{^nC_0} = \sum_{r=0}^n \frac{n-r}{^nC_r}$.Adding both expressions for $b_n$:$2b_n = \sum_{r=0}^n \frac{r + (n-r)}{^nC_r} = \sum_{r=0}^n \frac{n}{^nC_r} = n \sum_{r=0}^n \frac{1}{^nC_r}$.Since $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,we have $2b_n = na_n$.Therefore,$b_n = \frac{1}{2}na_n$.

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

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