## Calculus (3rd Edition)

$$\frac{35}{3}$$
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*}