Answer
$S_{1}=\dfrac {1}{3}$
$S_{2}=\dfrac {4}{9}$
$S_{3}=\dfrac {13}{27}$
$S_{4}=\dfrac {40}{81}$
$S_{5}=\dfrac {121}{243}$
$S_{6}=\dfrac {364}{729}$
Work Step by Step
This is geometric sequence
$S_{n}=\dfrac {a_{1}\left( 1-r^{n}\right) }{1-r}$
$S_{1}=\dfrac {\dfrac {1}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{1}\right) }{1-\dfrac {1}{3}}=\dfrac {1}{3}$
$S_{2}=\dfrac {\dfrac {1}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{2}\right) }{1-\dfrac {1}{3}}=\dfrac {4}{9}$
$S_{3}=\dfrac {\dfrac {1}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{3}\right) }{1-\dfrac {1}{3}}=\dfrac {13}{27}$
$S_{4}=\dfrac {\dfrac {1}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{4}\right) }{1-\dfrac {1}{3}}=\dfrac {40}{81}$
$S_{5}=\dfrac {\dfrac {1}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{5}\right) }{1-\dfrac {1}{3}}=\dfrac {121}{243}$
$S_{6}=\dfrac {\dfrac {1}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{6}\right) }{1-\dfrac {1}{3}}=\dfrac {364}{729}$
