#### Answer

$$\frac{63}{40}$$

#### Work Step by Step

Given
$$\sum_{n=2}^{n=\infty} \frac{7 \cdot(-3)^{n}}{5^{n}}=\sum_{n=2}^{n=\infty} 7\left(\frac{-3}{5}\right)^{n}$$
Since the series is a geometric series with $|r|=\left|\frac{3}{5}\right|<1$, then the series converges and has the sum
\begin{align*}
S&= \frac{a_1}{1-r}\\
&=\frac{ 7\left(\frac{-3}{5}\right)^2}{1-\frac{-3}{5}}\\
&= \frac{63}{40}
\end{align*}