## College Algebra 7th Edition

$S_{1}=\frac{1}{3}$ $S_{2}=\frac{4}{9}$ $S_{3}=\frac{13}{27}$ $S_{4}=\frac{40}{81}$ $S_{5}=\frac{121}{243}$ $S_{6}=\frac{364}{729}$
We are given: $a_{1}= \frac{1}{3}$ $a_{2}= \frac{1}{3^{2}}$ $a_{3}= \frac{1}{3^{3}}$ $a_{4}= \frac{1}{3^{4}}$ We notice that the denominator equals powers of $3$. So: $a_{n}= \frac{1}{3^{n}}$ Thus: $a_{5}= \frac{1}{3^{5}}$ $a_{6}= \frac{1}{3^{6}}$ We find the partial sums: $S_{1}= \frac{1}{3}$ $S_{2}= \frac{1}{3}+\frac{1}{3^{2}}=\frac{4}{9}$ $S_{3}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}=\frac{13}{27}$ $S_{4}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}=\frac{40}{81}$ $S_{5}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}=\frac{121}{243}$ $S_{6}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}+\frac{1}{3^{6}}=\frac{364}{729}$