Answer
$S_{n}=1-\left( \dfrac {1}{3}\right) ^{n}$
$\Rightarrow S_{1}=\dfrac {2}{3}$
$\Rightarrow S_{2}=\dfrac {8}{9}$
$\Rightarrow S_{3}=\dfrac {26}{27}$
$\Rightarrow S_{4}=\dfrac {80}{81}$
Work Step by Step
This is geometric sequence
$r=\dfrac {1}{3}$
$a_{1}=\dfrac {2}{3^{1}}=\dfrac {2}{3}$
$$S_{n}=\dfrac {a_{1}\left( 1-r^{n}\right) }{1-r}=\dfrac {\dfrac {2}{3}\left( 1-\left( \dfrac {1}{3}\right) ^{n}\right) }{1-\dfrac {1}{3}}=1-\left( \dfrac {1}{3}\right) ^{n}$$
$\Rightarrow S_{1}=1-\left( \dfrac {1}{3}\right) ^{1}=\dfrac {2}{3}$
$\Rightarrow S_{2}=1-\left( \dfrac {1}{3}\right) ^{2}=\dfrac {8}{9}$
$\Rightarrow S_{3}=1-\left( \dfrac {1}{3}\right) ^{3}=\dfrac {26}{27}$
$\Rightarrow S_{4}=1-\left( \dfrac {1}{3}\right) ^{4}=\dfrac {80}{81}$
