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

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(C) Let $b = \sum_{r=0}^n \frac{r}{{}^n C_r}$  ....$(i)$Replace $r$ by $n-r$:$b = \sum_{r=0}^n \frac{n-r}{{}^n C_{n-r}}$Since ${}^n C_r = {}^n C_{n-r}

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$\frac{n}{2} \cdot a_n$

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(C) Let $b = \sum_{r=0}^n \frac{r}{{}^n C_r}$  ....$(i)$Replace $r$ by $n-r$:$b = \sum_{r=0}^n \frac{n-r}{{}^n C_{n-r}}$Since ${}^n C_r = {}^n C_{n-r}$,we have:$b = \sum_{r=0}^n \frac{n-r}{{}^n C_r}$  ....$(ii)$Adding $(i)$ and $(ii)$:$2b = \sum_{r=0}^n \frac{r + n - r}{{}^n C_r} = \sum_{r=0}^n \frac{n}{{}^n C_r}$$2b = n \sum_{r=0}^n \frac{1}{{}^n C_r}$Since $a_n = \sum_{r=0}^n \frac{1}{{}^n C_r}$,we get:$2b = n \cdot a_n \Rightarrow b = \frac{n}{2} \cdot a_n$

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

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