College Algebra 7th Edition

$S_4 = \dfrac{80}{81}$
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: $\require{cancel} S_4 =\dfrac{2}{3}\left(\dfrac{1-(\frac{1}{3})^4}{1-\frac{1}{3}}\right) \\S_4=\dfrac{2}{3}\left(\dfrac{1-\frac{1}{81}}{\frac{2}{3}}\right) \\S_4=\dfrac{2}{3}\left(\dfrac{\frac{80}{81}}{\frac{2}{3}}\right) \\S_4=\dfrac{2}{3} \cdot \dfrac{80}{81} \cdot \dfrac{3}{2} \\S_4=\dfrac{\cancel{2}}{\cancel{3}} \cdot \dfrac{80}{81} \cdot \dfrac{\cancel{3}}{\cancel{2}} \\S_4 = \dfrac{80}{81}$