# How do you find the sum of the finite geometric sequence of sum_(i=1)^100 15(2/3)^(i-1)?

May 18, 2018

${\sum}_{i = 1}^{100} 15 {\left(\frac{2}{3}\right)}^{i - 1} =$

#### Explanation:

To find the sum ${\sum}_{k = 0}^{n} {a}_{0} {r}^{n}$ of a geometric series with first term ${a}_{0}$ and ratio $r$, we use the following formula

${\sum}_{k = 0}^{n} {a}_{0} {r}^{n} = \frac{{a}_{0} \left(1 - {r}^{n}\right)}{1 - r}$

So, if $k = i - 1$, then

${\sum}_{i = 1}^{100} 15 {\left(\frac{2}{3}\right)}^{i - 1} = 15 {\sum}_{k = 0}^{99} {\left(\frac{2}{3}\right)}^{k} = 15 \left(\frac{1 \left(1 - {\left(\frac{2}{3}\right)}^{99}\right)}{1 - \frac{2}{3}}\right) = 45$