K kilroymcb New member Joined Mar 25, 2007 Messages 27 Mar 29, 2007 #1 This asks me to find the sum of the series from n=1 to infinity of ((-3)^(n+2))/7^(n-2). I'm thinking this is going to be a geometric series, but I'm not entirely certain how to evaluate it. Any hints?
This asks me to find the sum of the series from n=1 to infinity of ((-3)^(n+2))/7^(n-2). I'm thinking this is going to be a geometric series, but I'm not entirely certain how to evaluate it. Any hints?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Mar 29, 2007 #2 We can simplify it down and then use a geometric series. \(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{(-3)^{n+2}}{7^{n-2}}=\sum_{n=1}^{\infty}\frac{9\cdot{(-3)^{n}}}{7^{n}\cdot{7^{-2}}}=441\sum_{n=1}^{\infty}\frac{(-3)^{n}}{7^{n}}\) We can make our series out of this sum. \(\displaystyle \L\\S=\sum_{n=1}^{\infty}\frac{(-3)^{n}}{7^{n}}=\frac{-3}{7}+\frac{9}{49}-\frac{27}{343}+\frac{81}{2401}-...................\) Factor out -3/7: \(\displaystyle \L\\S=\frac{-3}{7}(1\underbrace{-\frac{3}{7}+\frac{9}{49}-\frac{27}{343}+..........}_{\text{This is S}})\) So, we have: \(\displaystyle \L\\S=\frac{-3}{7}(1+S)\) \(\displaystyle \L\\S=\frac{-3}{10}\) Don't forget the 441 up there: \(\displaystyle \L\\S=\frac{-1323}{10}\)
We can simplify it down and then use a geometric series. \(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{(-3)^{n+2}}{7^{n-2}}=\sum_{n=1}^{\infty}\frac{9\cdot{(-3)^{n}}}{7^{n}\cdot{7^{-2}}}=441\sum_{n=1}^{\infty}\frac{(-3)^{n}}{7^{n}}\) We can make our series out of this sum. \(\displaystyle \L\\S=\sum_{n=1}^{\infty}\frac{(-3)^{n}}{7^{n}}=\frac{-3}{7}+\frac{9}{49}-\frac{27}{343}+\frac{81}{2401}-...................\) Factor out -3/7: \(\displaystyle \L\\S=\frac{-3}{7}(1\underbrace{-\frac{3}{7}+\frac{9}{49}-\frac{27}{343}+..........}_{\text{This is S}})\) So, we have: \(\displaystyle \L\\S=\frac{-3}{7}(1+S)\) \(\displaystyle \L\\S=\frac{-3}{10}\) Don't forget the 441 up there: \(\displaystyle \L\\S=\frac{-1323}{10}\)
