# How do you find the sum of the infinite geometric series given Sigma 48(2/3)^(n-1) from n=1 to oo?

Oct 25, 2016

$144$

#### Explanation:

First write out the GP, taking the $48$ out as a common factor ( to make the calculations easier)

$48 {\sum}_{1}^{\infty} {\left(\frac{2}{3}\right)}^{n - 1} = 48 \left(1 + \frac{2}{3} + {\left(\frac{2}{3}\right)}^{2} + {\left(\frac{2}{3}\right)}^{3} + \ldots . . +\right)$

for this is GP first term $a = 1$, common ratio $r = \frac{2}{3}$

the sum to infinity of GP is ${S}_{O O} = \frac{a}{1 - r}$

providing $| r | < 1$

in this case $r = \frac{2}{3} < 1 , : > {S}_{O O}$ exists

${S}_{O O} = \frac{a}{1 - r} = 48 \left(\frac{1}{1 - \left(\frac{2}{3}\right)}\right)$

$= 48 \left(\frac{1}{\frac{1}{3}}\right) = 48 \times 3 = 144$