# How do you find the sum of the finite geometric sequence of Sigma 3(3/2)^n from n=0 to 20?

Mar 28, 2018

${\sum}_{n = 0}^{20} 3 {\left(\frac{3}{2}\right)}^{n} \approx 29921.31057$

#### Explanation:

${\sum}_{n = 0}^{20} 3 {\left(\frac{3}{2}\right)}^{n} = 3 + 3 \left(\frac{3}{2}\right) + 3 {\left(\frac{3}{2}\right)}^{2} + 3 {\left(\frac{3}{2}\right)}^{3} + \cdots + 3 {\left(\frac{3}{2}\right)}^{20}$

This is a geometric series of the form

${\sum}_{n = 0}^{k} a {R}^{n}$

with $a = 3$ and $R = \frac{3}{2}$

There is a formula to calculate the $k$th partial sum of a geometric series. It is given by:

${S}_{k} = a \left(\frac{1 - {R}^{k + 1}}{1 - R}\right)$

Therefore:

${S}_{20} = a \left(\frac{1 - {R}^{21}}{1 - R}\right) = 3 \left(\frac{1 - {\left(\frac{3}{2}\right)}^{21}}{1 - \frac{3}{2}}\right)$

$= 3 \left(\frac{1 - {\left(\frac{3}{2}\right)}^{21}}{- \frac{1}{2}}\right) = - 6 \left(1 - {\left(\frac{3}{2}\right)}^{21}\right) = 6 \left({\left(\frac{3}{2}\right)}^{21} - 1\right)$

Obviously this is going to be a pretty big number. If you punch it into a calculator, you'll find it's

$\approx 29921.31057$