Explanation

(a)
The general term of the given series is
$T_{r} = ({- 1)}^{r}(3 + 5r)\ ^{n}C_{r}$
$\therefore Sum\ \sum_{r = 0}^{n}(1)^{r}(3 + 5r)\ ^{n}C_{r}$
$= 3\begin{matrix}
n \\
\ \\
r = 0
\end{matrix}({- 1)}^{r}\ ^{n}C_{r} + 5\sum_{r = 0}^{n}(1)^{r}\ r^{n}C_{r}$
$= 3(C_{0} - C_{1} + C_{2} - C_{3} + C_{4} -$…$+ ( - 1)^{n} \bullet C_{n})$
$+ 5\left( - C_{1} + 2C_{2} - 3C_{3} + 4C_{4} - \ldots + ( - 1)^{n} \bullet n \bullet C_{n} \right)$
$\Longrightarrow S = 0 + 0 = 0$