Answer
$S_{5}= 155$
Work Step by Step
Given $a_{n} = 5(2)^{n-1}$
It is in the form of geometric sequence $a_{n} =a_{1}(r)^{n-1}$
$a_{1} = 5$
$r=2$
Partial sum of geometric sequence $S_{n} = \frac{a_{1}(1-r^{n})}{1-r}$
Substituting $a_{1} ,r$ values and $n=5$
$S_{5} = \frac{5(1-2^{5})}{1-2}$
$S_{5} = \frac{5(1-32)}{-1}$
$S_{5} = \frac{5(-31)}{-1}$
$S_{5} = 155$
