#### Answer

$$\frac{8}{5}$$

#### Work Step by Step

Given $$\frac{2^{3}}{7}+\frac{2^{4}}{7^{2}}+\frac{2^{5}}{7^{3}}+\frac{2^{6}}{7^{4}}+\cdots=\sum_{n=0}^{n=\infty} \frac{8}{7}\left(\frac{2}{7}\right)^{n} $$
Since the series is a geometric series with $|r|= \frac{2}{7}<1$, then the series converges and has the sum
\begin{align*}
S &=\frac{a_1}{1-r} \\
&=\frac{\frac{8}{7}}{1-\frac{2}{7}} \\
&=\frac{8}{5}
\end{align*}