## College Algebra 7th Edition

$S_{1}=\frac{2}{3}$ $S_{2}=\frac{8}{9}$ $S_{3}=\frac{26}{27}$ $S_{4}=\frac{80}{81}$ $S_n=\frac{3^n-1}{3^n}$
We are given: $a_{n}= \frac{2}{3^{n}}$ We find the partial sums: $S_{1}= \frac{2}{3^1}$ $S_{2}= \frac{2}{3^1}+\frac{2}{3^{2}}=\frac{8}{9}$ $S_{3}= \frac{2}{3^1}+\frac{2}{3^{2}}+\frac{2}{3^{3}}=\frac{26}{27}$ $S_{4}= \frac{2}{3^1}+\frac{2}{3^{2}}+\frac{2}{3^{3}}+\frac{2}{3^{4}}=\frac{80}{81}$ We notice that the partial sums have powers of $3$ in the denominator and $1$ value less in the numerator. Therefore: $S_n=\frac{3^n-1}{3^n}$