# How do you find the sum of the finite geometric sequence of Sigma 5(-3/2)^(n-1) from n=1 to 8?

Jan 9, 2018

$- \frac{6305}{128}$

#### Explanation:

In general for finite geometric series:

${\sum}_{k = 1}^{n} \left(a \cdot {r}^{k - 1}\right) = \frac{a \cdot \left(1 - {r}^{n}\right)}{1 - r}$

In this case $a = 5$, $r = - \frac{3}{2}$, and $n = 8$, so:

${\sum}_{k = 1}^{n} \left(a \cdot {r}^{k - 1}\right) = \frac{5 \cdot \left(1 - {\left(- \frac{3}{2}\right)}^{8}\right)}{1 - \left(- \frac{3}{2}\right)} = - \frac{6305}{128}$