Geometric Series - Finding a Partial Sum Equation

Sean O
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Is it possible to find the partial sum equation for (2^m - 1)/3^m, from m=0 to m=n-1?

I know that I'm supposed to rearrange the expression into the format ar^m, so the exponent m must only be on the value r, and not on the constant a. So far the farthest I've gotten is to rearrange it into (2/3)^m - (1/3)^m, but I have no idea what to do from there. I'm also not sure how I would go about proving that this expression can't be made into a partial sum, if it turns out it isn't possible.

Any input or hints would be a big help.
 
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You can just sum up (2/3)^m and (1/3)^m separately with the usual formula, then subtract the results.
 
Damn, that's one of those "smack-yourself-for-not-noticing-it" solutions. :smile: That makes perfect sense. Thanks for your help.
 
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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