# How do you find sum of geometric series 2+6+18+...+1458?

Mar 15, 2016

$2186$

#### Explanation:

Given that first term ${a}_{1} = 2$

Common ratio $q = {a}_{2} / {a}_{1} = \frac{6}{2} = 3$

General expression for $n t h$ term is

${a}_{n} = {a}_{1} \cdot {q}^{n - 1}$

To ascertain what is number of the last term, insert in the above equation

$1458 = 2 \cdot {3}^{n - 1}$
$\implies {3}^{n - 1} = \frac{1458}{2} = 729$
Writing $729$ as power of $3$ we obtain

${3}^{n - 1} = {3}^{6}$, comparing the exponents
$n - 1 = 6$
or $n = 7$
Now sum of $n$ terms is given by the expression

${S}_{n} = \frac{{a}_{1} \left(1 - {q}^{n}\right)}{1 - q}$

${S}_{7} = \frac{2 \left(1 - {3}^{7}\right)}{1 - 3}$
$= \frac{2 \left(1 - 2187\right)}{-} 2 = 2186$