Let an=∑r=0n 1 nCr2 and bn=∑r=0n (n−r) nCr2then bn equals

We havebn=∑r=0n n−r nCr2=∑r=0n n−r nCn−r2=∑r=0n r nCr2Thus, bn+bn=∑r=0n n−rnCr2+∑r=0n rnCr2=n∑r=0n 1 nCr2=nan⇒ bn=n2an

Let an=∑r=0n 1 nCr2 and bn=∑r=0n (n−r) nCr2then bn equals

We havebn=∑r=0n n−r nCr2=∑r=0n n−r nCn−r2=∑r=0n r nCr2Thus, bn+bn=∑r=0n n−rnCr2+∑r=0n rnCr2=n∑r=0n 1 nCr2=nan⇒ bn=n2an

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Let $a_{n} = \sum_{r = 0}^{n} \frac{1}{{(^{n} C_{r})}^{2}}$ and $b_{n} = \sum_{r = 0}^{n} \frac{(n - r)}{{(^{n} C_{r})}^{2}}$
then $b_{n}$ equals

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$a_{n}$

$n a_{n}$

$\frac{n}{2} a_{n}$

answer is C.

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Detailed Solution

We have

$b_{n} = \sum_{r = 0}^{n} \frac{n - r}{{(^{n} C_{r})}^{2}} = \sum_{r = 0}^{n} \frac{n - r}{{(^{n} C_{n - r})}^{2}} = \sum_{r = 0}^{n} \frac{r}{{(^{n} C_{r})}^{2}}$

Thus,

$b_{n} + b_{n} = \sum_{r = 0}^{n} \frac{n - r}{{(n_{C_{r}})}^{2}} + \sum_{r = 0}^{n} \frac{r}{{(n_{C_{r}})}^{2}} = n \sum_{r = 0}^{n} \frac{1}{{(^{n} C_{r})}^{2}} = n a_{n}$