Answer
$a_1=\frac{1}{2}$
$a_2=\frac{3}{4}$
$a_3=\frac{7}{8}$
$a_4=\frac{15}{16}$
Work Step by Step
We know that $a_n=\frac{2^n-1}{2^n}$. Hence, here:
$a_1=\frac{2^1-1}{2^1}=\frac{2-1}{2}=\frac{1}{2}$
$a_2=\frac{2^2-1}{2^2}=\frac{4-1}{4}=\frac{3}{4}$
$a_3=\frac{2^3-1}{2^3}=\frac{8-1}{8}=\frac{7}{8}$
$a_4=\frac{2^4-1}{2^4}=\frac{16-1}{16}=\frac{15}{16}$
