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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

a_{n}

b

n a_{n}

c

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

d

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detailed solution

Correct option is C

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,

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

$\Rightarrow b_{n} = \frac{n}{2} a_{n}$

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