Explanation

(a)
We have, $\sum_{r = 0}^{n}{\sum_{s = 0}^{n}{(r + s)(C_{r} + C_{s})}}$
$= \sum_{r = 0}^{n}{\sum_{s = 0}^{n}\left( rC_{r} + rC_{s} + sC_{r} + sC_{s} \right)}$
$= \sum_{r = 0}^{n}\left\lbrack \sum_{s = 0}^{n}{rC_{r} + r\sum_{s = 0}^{n}{C_{s} + C_{s}\sum_{s = 0}^{n}{s + \sum_{s = 0}^{n}{sC_{s}}}}} \right\rbrack$
$= \sum_{r = 0}^{n}\left\lbrack (n + 1)r \bullet C_{r} + r2^{n} + \frac{n(n + 1)}{2}C_{r} + n \bullet 2^{n - 1} \right\rbrack$
$= (n + 1)n \bullet 2^{n - 1} + \left( 2^{n} \right)\frac{n(n + 1)}{2} + \frac{n(n + 1)}{2}2^{n} + n2^{n - 1}(n + 1)$
$= n(n + 1)2^{n} + n(n + 1)2^{n}$
$= 2n(n + 1)2^{n}$ …(i)
Also, $\sum_{r = 0}^{n}{\sum_{s = 0}^{n}{(r + s)\left( C_{r} + C_{s} \right)}}$
$= \sum_{r = 0}^{n}{4rC_{r} + 2}\sum_{0 \leq r < s \leq n}^{}{\sum_{}^{}{(r + s)(C_{r} + C_{s})}}$
$\therefore\ 2n(n + 1)2^{n} = 4n \bullet 2^{n - 1} + 2\sum_{0 \leq r < s \leq n}^{}{\sum_{}^{}{(r + s)(C_{r} + C_{s})}}$
$\Rightarrow \ \sum_{0 \leq r < s \leq n}^{}{\sum_{}^{}{(r + s)(C_{r} + C_{s})}} = n^{2} \bullet 2^{n}$