# How do you find the sum of the finite geometric sequence of Sigma 5(3/5)^n from n=0 to 40?

Oct 13, 2017

12.5 (1.d.p)

#### Explanation:

First calculate the first three terms:

$5 {\left(\frac{3}{5}\right)}^{0} = \textcolor{b l u e}{5} , 5 {\left(\frac{3}{5}\right)}^{1} = \textcolor{b l u e}{3} , 5 {\left(\frac{3}{5}\right)}^{2} = \textcolor{b l u e}{\frac{9}{5}}$

Find the common ratio:

$\frac{3}{5} = \frac{\frac{9}{5}}{3} = \frac{3}{5}$

The sum of a geometric sequence is:

$a \left(\frac{1 - {r}^{n}}{1 - r}\right)$

Where a is the first term, n is the nth term and r is the common ratio.

So:

$5 \left(\frac{1 - {\left(\frac{3}{5}\right)}^{40}}{1 - \left(\frac{3}{5}\right)}\right) = 12.4999999899743791$(16 .d.p.)