Answer
$a_{n}=\frac{(-1)^{n}}{3^{n}}$
Work Step by Step
We are given:
$-\frac{1}{3}, \frac{1}{9}, - \frac{1}{27}, \frac{1}{81}$, ...
We notice that there are powers of 3 in the denominator. In addition, the signs of the terms alternate. We find the pattern:
$a_{1}=\frac{(-1)^{1}}{3^{1}}$
$a_{2}=\frac{(-1)^{2}}{3^{2}}$
$a_{3}=\frac{(-1)^{3}}{3^{3}}$
$a_{4}=\frac{(-1)^{4}}{3^{4}}$
...
$a_{n}=\frac{(-1)^{n}}{3^{n}}$
