# How do you find the sum of the finite geometric sequence of Sigma 2(-1/4)^n from n=0 to 40?

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Roy E. Share
Mar 8, 2018

$\left(\frac{8}{5}\right) \cdot \left(1 + {\left(\frac{1}{4}\right)}^{41}\right)$
$\cong 1.60000000 \ldots$

#### Explanation:

There are 41 terms in the series. The common ratio, $r$ is $- \left(\frac{1}{4}\right)$ and the first term, $a$ is $2 \left(- \frac{1}{4}\right) 60$ which is $2$. Therefore the sum of the first 41 terms is $a \cdot \frac{1 - {r}^{n}}{1 - r}$
$= 2 \cdot \frac{1 - {\left(- \frac{1}{4}\right)}^{41}}{1 - \left(- \frac{1}{4}\right)}$
$= \frac{8}{5} \cdot \left(1 + {\left(\frac{1}{4}\right)}^{41}\right)$ because 41 is odd
$\cong 1.6$
because ${\left(\frac{1}{4}\right)}^{41}$ is negligible

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