by looking at the expression, we can pretty much see who's the common factor and first term.
[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}
\end{cases}
\\\\[-0.35em]
\rule{34em}{0.25pt}[/tex]
[tex]\bf \sum\limits_{n=1}^{4}~(\stackrel{\stackrel{a_1}{\downarrow }}{-2})(\stackrel{\stackrel{r}{\downarrow }}{-3})^{n-1}\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\[-0.5em]
\hrulefill\\
a_1=-2\\
r=-3\\
n=4
\end{cases}
\\\\\\
S_4=-2\left( \cfrac{1-(-3)^4}{1-(-3)} \right)\implies S_4=-2\left( \cfrac{1-81}{1+3} \right)
\\\\\\
S_4=-2\cdot \cfrac{-80}{4}\implies S_4=-\cfrac{-80}{2}\implies S_4=40[/tex]
