# How do you find the sum of the infinite geometric series (1 - x) + (x^3 – x^4) + (x^6 – x^7) + ….. + [x^(3n) - x^(3n+1)] ?

Nov 16, 2015

${\sum}_{n = 0}^{\infty} \left({x}^{3 n} - {x}^{3 n + 1}\right) = \frac{1}{1 + x + {x}^{2}}$

with radius of convergence $1$

#### Explanation:

${\sum}_{n = 0}^{\infty} \left({x}^{3 n} - {x}^{3 n + 1}\right)$

$= \left(1 - x\right) {\sum}_{n = 0}^{\infty} {x}^{3 n}$

$= \left(1 - x\right) {\sum}_{n = 0}^{\infty} {\left({x}^{3}\right)}^{n}$

$= \frac{1 - x}{1 - {x}^{3}} \left(1 - {x}^{3}\right) {\sum}_{n = 0}^{\infty} {\left({x}^{3}\right)}^{n}$

$= \frac{1 - x}{1 - {x}^{3}} \left({\sum}_{n = 0}^{\infty} {\left({x}^{3}\right)}^{n} - {x}^{3} {\sum}_{n = 0}^{\infty} {\left({x}^{3}\right)}^{n}\right)$

$= \frac{1 - x}{1 - {x}^{3}} \left({\sum}_{n = 0}^{\infty} {\left({x}^{3}\right)}^{n} - {\sum}_{n = 1}^{\infty} {\left({x}^{3}\right)}^{n}\right)$

$= \frac{1 - x}{1 - {x}^{3}}$

$= \frac{1 - x}{\left(1 - x\right) \left(1 + x + {x}^{2}\right)}$

$= \frac{1}{1 + x + {x}^{2}}$

with radius of convergence $1$