Answer
$\displaystyle \sum_{j=0}^{40}2(-\frac{1}{4})^n\approx1.6$
Work Step by Step
$\displaystyle \sum_{j=0}^{40}2(-\frac{1}{4})^n=2(-\frac{1}{4})^0+2(-\frac{1}{4})^1+2(-\frac{1}{4})^2+...+2(-\frac{1}{4})^{40}$
There are 41 terms is this sequence.
$a_1=2(-\frac{1}{4})^0=2$
$r=\frac{a_2}{a_1}=\frac{2(-\frac{1}{4})^1}{2(-\frac{1}{4})^0}=-\frac{1}{4}$
$S_n=a_1(\frac{1-r^n}{1-r})$
$S_{41}=2(\frac{1-(-\frac{1}{4})^{41}}{1-(-\frac{1}{4})})=2(\frac{1-(-\frac{1}{4})^{41}}{\frac{5}{4}})\approx1.6$
