Answer:
Option D
Explanation:
The general term of the given series is
$T_{r}=(-1)^{r}(3+5r)^{n}C_{r},r=0,1,2,.....,n$
$\therefore$ $S_{n}=\sum_{r=0}^{n}(-1)^{r}(3+5r)^{n} C_{r}$
$=3\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}+5\sum_{r=0}^{n}(-1)^{r}.{r^{n}}C_{r}$
$=3\left[\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}\right]+5\left[\sum_{r=1}^{n}(-1)^{r}.{r^{}\frac{n}{r}}.{^{n-1}}C_{r-1}\right]$
$=3\left[\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}\right]+5n\left[\sum_{r=1}^{n}(-1)^{r}.{^{n-1}}C_{r-1}\right]$
=$3(0)+5n(0)=0+0=0$
