Answer
The sum of the series $S_{n}$ = 8
Work Step by Step
The given infinite geometric series = 4 + $\frac{4}{2}$ + $\frac{4}{2^{2}}$ + $\frac{4}{2^{3}}$ ........................
The common ratio = $\frac{\frac{4}{2^{3}}}{\frac{4}{2^{2}}}$ = $\frac{1}{2}$
The first term $a_{1}$ = 4
The sum of the series $S_{n}$ = $\frac{a_{1}}{1 - r}$ = $\frac{4}{1 - (\frac{1}{2})}$ = $\frac{4}{\frac{1}{2}}$ = 4$\times$2 = 8
