## Algebra and Trigonometry 10th Edition

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