Answer
$$\frac{35}{3}$$
Work Step by Step
Given $$\sum_{n=0}^{n=\infty} \frac{8+2^{n}}{5^{n}}=\sum_{n=0}^{n=\infty} 8 \cdot\left(\frac{1}{5}\right)^{n}+\sum_{n=0}^{n=\infty}\left(\frac{2}{5}\right)^{n}$$
Since the series is a geometric series with $|r_1|= \frac{1}{5}<1$ and $|r_2|= \frac{2}{5}<1$, then the series converges and has the sum
\begin{align*}
S&=S_1+S_2\\
&=\frac{a_1}{1-r_1}+\frac{a_1}{1-r_2}\\
&=\frac{8}{1-\frac{1}{5}}+\frac{1}{1-\frac{2}{5}}\\
&= \frac{35}{3}
\end{align*}
