Answer
$S_{21}=29,921.31$
Work Step by Step
$\displaystyle \sum_{n=0}^{20}3(\frac{3}{2})^n=3(\frac{3}{2})^0+3(\frac{3}{2})^1+3(\frac{3}{2})^2+...+3(\frac{3}{2})^{20}$
$a_1=3(\frac{3}{2})^0=3$
$r=\frac{a_2}{a_1}=\frac{3(\frac{3}{2})^1}{3(\frac{3}{2})^0}=\frac{3}{2}$
Notice that, from $0$ to $20$ there are 21 terms.
$S_{21}=a_1(\frac{1-r^n}{1-r})=3(\frac{1-(\frac{3}{2})^{21}}{1-\frac{3}{2}})=3·\frac{(\frac{3}{2})^{21}-1}{\frac{1}{2}}=
6·(1.5^{21}-1)=29,921.31$
