## Algebra 2 (1st Edition)

$\displaystyle \sum_{n=1}^{4}(\frac{1}{3^{n}})\qquad$or $\displaystyle \qquad\sum_{n=1}^{4}3^{-n}$
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 $\displaystyle \qquad\sum_{n=1}^{4}3^{-n}$