Answer
$\displaystyle\sum_{i=1}^{4}\left[12\left( \dfrac{1}{3} \right)^{i-1}\right]$
Work Step by Step
Using the summation notation, the series $
12+4+\dfrac{4}{3}+\dfrac{4}{9}
$ is equivalent to
\begin{array}{l}
\displaystyle\sum_{i=1}^{4}\left[12\left( \dfrac{1}{3} \right)^{i-1}\right]
.\end{array}