Explanation

(a)
Given, $s_{n} = \sum_{r = 0}^{n}\frac{1}{\ ^{n}C_{r}} = \sum_{r = 0}^{n}\frac{1}{\ ^{n}C_{r}}$ $\lbrack\because\ \ ^{n}C_{r} = \ ^{n}C_{r}\rbrack$
$\Rightarrow ns_{n} = \sum_{r = 0}^{n}\frac{n}{\ ^{n}C_{n - r}} = \sum_{r = 0}^{n}\left\lbrack \frac{n - r}{\ ^{n}C_{n - r}} + \frac{r}{\ ^{n}C_{n - r}} \right\rbrack$
$\Rightarrow ns_{n} = \sum_{r = 0}^{n}\frac{n - r}{\ ^{n}C_{n - r}} + \sum_{r = 0}^{n}\frac{r}{\ ^{n}C_{r}}$
$\Rightarrow \ \ \ ns_{n} = \left( \frac{n}{{\ ^{n}C}_{n}} + \frac{n - 1}{{\ ^{n}C}_{n - 1}} + ... + \frac{1}{{\ ^{n}C}_{0}} \right)\sum_{r = 0}^{n}\frac{r}{\ ^{n}C_{r}}$
$\Rightarrow \ \ \ ns_{n} = t_{n} + t_{n} = 2t_{n}$
$\Rightarrow \ \ \ \ \ \ \ \frac{t_{n}}{s_{n}} = \frac{n}{2}$