Answer
$S_{1}=\frac{1}{3}$
$S_{2}=\frac{4}{9}$
$S_{3}=\frac{13}{27}$
$S_{4}=\frac{40}{81}$
$S_{5}=\frac{121}{243}$
$S_{6}=\frac{364}{729}$
Work Step by Step
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:
$S_{1}= \frac{1}{3}$
$S_{2}= \frac{1}{3}+\frac{1}{3^{2}}=\frac{4}{9}$
$S_{3}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}=\frac{13}{27}$
$S_{4}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}=\frac{40}{81}$
$S_{5}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}=\frac{121}{243}$
$S_{6}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}+\frac{1}{3^{6}}=\frac{364}{729}$
