Answer
$a_n=\frac{1}{2^n}$
Work Step by Step
If we look at the pattern:
$ \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, ...$
$\frac{1}{2^1}, \frac{1}{2^2}, \frac{1}{2^3}, \frac{1}{2^4}, ...$
The nth term in the sequence is:
$a_n=\frac{1}{2^n}$
