Answer
The Sum of the given infinite geometric series = $\frac{4}{3}$
Work Step by Step
The Sum of a infinite geometric series ( if | r | $\lt$ 1 ) is given by
S = $\frac{First term }{1 - common ratio}$ = $\frac{a_{1}}{1 - r}$
The given infinite geometric series
= 1 + $\frac{1}{4}$ + $\frac{1}{16}$ + $\frac{1}{64}$ + ......................
Here
First term $a_{1}$ = 1
common ratio r = $\frac{\frac{1}{64}}{\frac{1}{16}}$ = $\frac{\frac{1}{16}}{\frac{1}{4}}$ = $\frac{\frac{1}{4}}{1}$ = $\frac{1}{4}$
The Sum of the given infinite geometric series = $\frac{a_{1}}{1 - r}$ = $\frac{1}{1 - \frac{1}{4}}$ = $\frac{1}{\frac{3}{4}}$ = $\frac{4}{3}$
