Computing a Geometric Power Series with Cosine Terms

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Homework Statement



Compute \sum_{n=0}^{\infty} p^n cos(3nx) for \abs{p} \textless 1, where p \in \mathbb{R}.

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The Attempt at a Solution



I was thinking that maybe this could be approached as a telescoping series, but I'm not really sure if it is. Would that be the most expedient approach? Clearly it isn't geometric, and I'm not sure how to find the sum of a general power series.
 
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No one? Someone has to know. There are a lot of clever people on this forum.
 
What if I rearrange it using the fact that cos(3nx) = 1/2(exp(3nix)-exp(-3inx)). Could I then write \sum_{n=0}^{\infty} (pe^{3xi})^n + \sum_{n=0}^\infty (pe^{-3ix})^n and try to work from there? (ie, is that valid?)
 
Ok, I was being silly, as usual - it is actually geometric after all.

Thank you to me for figuring out this problem.
 
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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