How Can You Evaluate This Infinite Series with a Recurrence Relation?

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Homework Statement



The question is to evaluate the infinite series of the Sum[(((-1)^n)*a(n))/10^n], as n goes from zero to infinity, and a(n) is the recurrence relation a(n)=5a(n-1)-6a(n-2) where a(0)=0, and a(1)=1


Homework Equations



I found the explicit equation for a(n)=3^n - 2^n, but I can't find how that will help. It doesn't really simplify the sum that I can tell.

The Attempt at a Solution



I think that if I could find a generating function for the recurrence relation, then it would probably be a lot easier to relate the series to something that I already know, but I am not sure how to find the generating function. Any help is much appreciated. Thanks a lot.
 
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Since you know that an= 3n- 2n and obvious thing to do is to put it in the sum- it certainly DOES simplify it!
The sum becomes
\sum_{n=}^{\infty}\frac{(-1)^n(3^n- 2^n)}{10^n}= \sum_{n=0}^\infty\left(\frac{-3}{10}\right)^n}-\sum_{n=0}^\infty\left(\frac{-2}{10}\right)^n
both of which are geometric series.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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