Explanation

(c)
Given, $a_{n} = \sum_{r = 0}^{n}\frac{1}{\ ^{n}C_{r}}$
Let $b_{n} = \sum_{r = 0}^{n}\frac{r}{\ ^{n}C_{r}}$
Then, $b_{n} = \frac{0}{\ ^{n}C_{0}} + \frac{1}{\ ^{n}C_{1}} + \frac{2}{\ \ ^{n}C_{2}} + .\ldots... + \frac{n}{\ ^{n}C_{n}}$ …(i)
$\Rightarrow b_{n} = \frac{n}{\ ^{n}C_{0}} + \frac{n - 1}{\ ^{n}C_{1}} + \frac{n - 2}{\ \ ^{n}C_{2}} + .\ldots.. + \frac{0}{\ ^{n}C_{n}}$ …(ii)
$\lbrack\because\ ^{n}C_{0} =^{n}C_{n},\ ^{n}C_{1} = \ ^{n}C_{n - 1}\ldots.as\ ^{n}C_{r} = \ ^{n}C_{n - r}\rbrack$
On adding Eqs. (i) and (ii), we get
${2b}_{n} = \frac{n}{\ ^{n}C_{0}} + \frac{n}{\ ^{n}C_{1}} + .\ldots.. + \frac{n}{\ ^{n}C_{n}}$
$= n\left\lbrack \frac{1}{\ ^{n}C_{0}} + \frac{1}{\ ^{n}C_{1}} + \frac{1}{\ \ ^{n}C_{2}} + .\ldots.. + \frac{1}{\ ^{n}C_{n}} \right\rbrack$
$\Rightarrow 2b_{n} = na_{n}$
$\therefore\ b_{n} = \frac{1}{2}na_{n}$