# How do you find the sum of the finite geometric sequence of Sigma 2(4/3)^n from n=0 to 15?

Jan 22, 2018

${\sum}_{n = 0}^{15} = 2 {\left(\frac{4}{3}\right)}^{n}$

$2 \left[{\left(\frac{4}{3}\right)}^{o} + {\left(\frac{4}{3}\right)}^{1} + {\left(\frac{4}{3}\right)}^{2.} . . {\left(\frac{4}{3}\right)}^{15}\right]$

Sum of any geometric finite geometric series is,

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

where,
$a =$ first term of the series
$r =$ common ratio of the series

substituting the values,

$2 \left[\frac{1 \left({\left(\frac{4}{3}\right)}^{16} - 1\right)}{\frac{4}{3} - 1}\right]$

$2 \left[\frac{\frac{{4}^{16} - {3}^{16}}{3} ^ 16}{\frac{1}{3}}\right]$

$2 \left[\frac{{4}^{15} - {3}^{15}}{3} ^ 15\right]$

Jan 22, 2018

${\sum}_{n = 0}^{15} 2 {\left(\frac{4}{3}\right)}^{n} \approx 592.647311$

#### Explanation:

Sum $= {\sum}_{n = 0}^{15} 2 {\left(\frac{4}{3}\right)}^{n} = 2 {\sum}_{n = 0}^{15} {\left(\frac{4}{3}\right)}^{n}$

Now, ${\sum}_{n = 0}^{15} {\left(\frac{4}{3}\right)}^{n}$ is the sum of geometric progression, with 1st term $a = 1$ and common ratio $r = \frac{4}{3}$

The sum of a finite GP $= \frac{a \left(1 - {r}^{n}\right)}{1 - r}$ where $n$ is the number of terms.

In our example, $a = 1 , r = \frac{4}{3} , n = 16$

$\therefore {\sum}_{n = 0}^{15} {\left(\frac{4}{3}\right)}^{n} = \frac{1 \cdot \left(1 - {\left(\frac{4}{3}\right)}^{16}\right)}{1 - \frac{4}{3}}$

$\approx - 98.77455184 \times - 3$

$\approx 296.3236555$

Sum $= 2 \cdot {\sum}_{n = 0}^{15} {\left(\frac{4}{3}\right)}^{n}$

$\approx 2 \times 296.3236555$

$\approx 592.647311$