Answer
$S_4 = \dfrac{80}{81}$
Work Step by Step
RECALL:
The partial sum $S_n$ (sum of the first $n$ terms) of a geometric sequence is given by the formula:
$S_n=a\left(\dfrac{1-r^n}{1-r}\right), r\ne 1$
where
$a$ = first term
$r$ = common ratio
The given geometric sequence has:
$a=\dfrac{2}{3}
\\r=\dfrac{1}{3}$
Thus, the sum of the first 4 terms is equal to:
$\require{cancel}
S_4 =\dfrac{2}{3}\left(\dfrac{1-(\frac{1}{3})^4}{1-\frac{1}{3}}\right)
\\S_4=\dfrac{2}{3}\left(\dfrac{1-\frac{1}{81}}{\frac{2}{3}}\right)
\\S_4=\dfrac{2}{3}\left(\dfrac{\frac{80}{81}}{\frac{2}{3}}\right)
\\S_4=\dfrac{2}{3} \cdot \dfrac{80}{81} \cdot \dfrac{3}{2}
\\S_4=\dfrac{\cancel{2}}{\cancel{3}} \cdot \dfrac{80}{81} \cdot \dfrac{\cancel{3}}{\cancel{2}}
\\S_4 = \dfrac{80}{81}$
