Answer
See below.
Work Step by Step
$a_n=\frac{1}{2}\cdot(\frac{1}{2})^{n-1}$ thus
$a_1=\frac{1}{2}\cdot(\frac{1}{2})^{1-1}=\frac{1}{2}\cdot1=\frac{1}{2}$
$a_2=\frac{1}{2}\cdot(\frac{1}{2})^{2-1}=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$
$a_3=\frac{1}{2}\cdot(\frac{1}{2})^{3-1}=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$
$a_4=\frac{1}{2}\cdot(\frac{1}{2})^{4-1}=\frac{1}{2}\cdot\frac{1}{8}=\frac{1}{16}$
$a_5=\frac{1}{2}\cdot(\frac{1}{2})^{5-1}=\frac{1}{2}\cdot\frac{1}{16}=\frac{1}{32}$
