Answer
$a_1 = \frac{1}{3}
\\a_2 = \frac{1}{9}
\\a_3 = \frac{1}{27}
\\a_4 = \frac{1}{81}$
Work Step by Step
To find the first four terms of the given sequence, substitute 1, 2, 3, and 4 to $n$, respectively, then simplify.
$a_1 = (\frac{1}{3})^1 = \frac{1}{3}
\\a_2 = (\frac{1}{3})^2 = (\frac{1}{3})(\frac{1}{3}) = \frac{1}{9}
\\a_3 = (\frac{1}{3})^3 = (\frac{1}{3})(\frac{1}{3})(\frac{1}{3}) = \frac{1}{27}
\\a_4 = (\frac{1}{3})^4 = (\frac{1}{3})(\frac{1}{3})(\frac{1}{3})(\frac{1}{3}) = \frac{1}{81}$
