sum of series from n=1 to infinity of ((-3)^(n+2))/7^(n-2)

[kilroymcb]

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?

 

[galactus]

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:

\(\displaystyle \L\\S=\frac{-1323}{10}\)