R Rostron New member Joined Aug 13, 2007 Messages 3 Aug 13, 2007 #1 I am having some trouble finding the sum of an infinite geometric series problem. Any help is appreciated! If x=2, find the sum of the series. \(\displaystyle \infty\) \(\displaystyle \Sigma\) \(\displaystyle \frac {(-1)^n(2x-3)^n}{4^n}\) n = 0
I am having some trouble finding the sum of an infinite geometric series problem. Any help is appreciated! If x=2, find the sum of the series. \(\displaystyle \infty\) \(\displaystyle \Sigma\) \(\displaystyle \frac {(-1)^n(2x-3)^n}{4^n}\) n = 0
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Aug 13, 2007 #2 \(\displaystyle \L x = 2\quad \Rightarrow \quad \sum\limits_{n = 0}^\infty {\left( {\frac{{ - 1}}{4}} \right)^n }.\) That is just a geometric series.
\(\displaystyle \L x = 2\quad \Rightarrow \quad \sum\limits_{n = 0}^\infty {\left( {\frac{{ - 1}}{4}} \right)^n }.\) That is just a geometric series.
