Answer
-$\frac{1}{3}$, $\frac{1}{9}$, -$\frac{1}{27}$ , $\frac{1}{81}$
Work Step by Step
To find the first four terms of the sequence whose general term is $a_{n}$ = $(\frac{1}{-3})^{n}$, we replace n in the formula with 1,2,3, and 4.
n=1, $a_{1}$ = $(\frac{1}{-3})^{1}$ = -$\frac{1}{3}$
n=2, $a_{2}$ =$(\frac{1}{-3})^{2}$ = $\frac{1}{-3}$* $\frac{1}{-3}$ = $\frac{1}{9}$
n=3, $a_{3}$ = $(\frac{1}{-3})^{3}$ = $\frac{1}{-3}$*$\frac{1}{-3}$* $\frac{1}{-3}$ = -$\frac{1}{27}$
n=4,$a_{4}$ =$(\frac{1}{-3})^{4}$ = $\frac{1}{-3}$*$\frac{1}{-3}$* $\frac{1}{-3}$* $\frac{1}{-3}$ = $\frac{1}{81}$
The first four terms are -$\frac{1}{3}$, $\frac{1}{9}$, -$\frac{1}{27}$ and $\frac{1}{81}$.
