Answer
$\displaystyle \sum_{n=1}^{4}(\frac{1}{3^{n}})\qquad $or $\displaystyle \qquad\sum_{n=1}^{4}3^{-n}$
Work Step by Step
There are four terms, ( $\displaystyle \sum_{n=1}^{4}$... )
in which the denominators are the first four powers of 3... $a_{n}=3^{n}$
Sum = $\displaystyle \sum_{n=1}^{4}(\frac{1}{3^{n}})\qquad $or $\displaystyle \qquad\sum_{n=1}^{4}3^{-n}$
