Answer
$\displaystyle \frac{45}{7}$
Work Step by Step
$\displaystyle \sum_{n=0}^{\infty}[(0.3)^{n}+(0.8)^{n}]=\sum_{n=0}^{\infty}(0.3)^{n}+\sum_{n=0}^{\infty}(0.8)^{n}$
Both are geometric series,
the first with $a=1, r=0.3 < 1$
the second with $a=1, r=0.8 < 1$.
(both converge by Th 9/6, $\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
$\displaystyle \sum_{n=0}^{\infty}[(0.3)^{n}+(0.8)^{n}]==\frac{1}{1-0.3}+\frac{1}{1-0.8}$
$=\displaystyle \frac{1}{0.7}+\frac{1}{0.2}$
$=\displaystyle \frac{10}{7}+5$
$=\displaystyle \frac{45}{7}$
