Answer
The sum of all terms = $\frac{5}{7}$
Work Step by Step
Put i = 1 to infinity in $(-\frac{2}{5})^{i - 1}$
First term (i = 1) $a_{1}$ = $(-\frac{2}{5})^{1- 1}$ = 1
Second term (i = 2) $a_{2}$ = $(-\frac{2}{5})^{2-1}$ = $-\frac{2}{5}$
Third term (i = 3) $a_{3}$ = $(-\frac{2}{5})^{3-1}$ = $\frac{4}{25}$
Fourth term (i = 4) $a_{4}$ = $(-\frac{2}{5})^{4-1}$ = $-\frac{8}{125}$
The sequence become = 1, $-\frac{2}{5}$, $\frac{4}{25}$, $-\frac{8}{125}$.......................
From above we observe
Common ratio = $-\frac{2}{5}$
The sum of all terms = $\frac{a_{1}}{1 - r}$
= $\frac{1}{1 -(-\frac{2}{5})}$ = $\frac{1}{1 + \frac{2}{5}}$= $\frac{1}{\frac{7}{5}}$ = $\frac{5}{7}$
