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

A
$a_{n}$
B
$n a_{n}$
C
$\frac{n}{2} a_{n}$
D
0

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,

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

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

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

Solution:

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