Answer
$S_{1}=\dfrac {3}{1}=3;$
$S_{2}=S_{1}+\dfrac {3}{2}=\dfrac {9}{2};$
$S_{3}=S_{2}+\dfrac {3}{4}=\dfrac {21}{4};$
$S_{4}=S_{3}+\dfrac {3}{8}=\dfrac {45}{8};$
$S_{5}=S_4+\frac{3}{16}=\frac{93}{16}$
Work Step by Step
$S_{n}=\sum ^{\infty }_{n=1}\dfrac {3}{2^{n-1}}\Rightarrow $
$S_{1}=\dfrac {3}{1}=3;$
$S_{2}=S_{1}+\dfrac {3}{2}=\dfrac {9}{2};$
$S_{3}=S_{2}+\dfrac {3}{4}=\dfrac {21}{4};$
$S_{4}=S_{3}+\dfrac {3}{8}=\dfrac {45}{8};$
$S_{5}=S_4+\frac{3}{16}=\frac{93}{16}$
