Answer
$S=12$
Work Step by Step
$\displaystyle \sum_{k=1}^{∞}4(\frac{2}{3})^{k-1}=4(\frac{2}{3})^{0}+4(\frac{2}{3})^{1}+4(\frac{2}{3})^{2}...$
$a_1=4(\frac{2}{3})^{0}=4$
$r=\frac{a_2}{a_1}=\frac{4(\frac{2}{3})^{1}}{4(\frac{2}{3})^{0}}=\frac{2}{3}$
$S=\frac{a_1}{1-r}=\frac{4}{1-\frac{2}{3}}=\frac{4}{\frac{1}{3}}=12$
