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$If a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}, then \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ is equal to

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(n−1)an

nan

12nan

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Correct option is C

Let b=∑r=0n r nCr=∑r=0n n−(n−r) nCr=n∑r=0n 1 nCr−∑r=0n n−r nCr=nan−∑r=0n n−r nCn−r ∵nCr=nCn−r=nan−b⇒2b=nan⇒b=n2an

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If an=∑r=0n 1 nCr, then ∑r=0n r nCr is equal to

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$If a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}, then \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ is equal to