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Value of $\sum x^n$

I was wondering how to derive a closed formula for things like $\sum_{i=1}^{n}2^{n}$=$2(2^{n}-1)$ and $\sum_{n=k}^{n}2^{n-k}$=$2^{n-k+1}-1$. I haven't done this in a while, and had wolfram do it for me, and I am not sure what the general tactic in getting these formulas is. Your help is appreciated!

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marked as duplicate by Aryabhata, Ross Millikan, Mike Spivey, Akhil Mathew Apr 6 '11 at 4:56

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Look up "geometric series". –  Robert Israel Apr 6 '11 at 4:40
    
But there the other two ways are not mentioned; thought they are essentially the same. –  awllower Apr 6 '11 at 4:43
    
I am still not clear on how to get something that starts not with 0 but with n-k for example, as in my example. –  user9185 Apr 6 '11 at 4:47
    
If you're seeking tricks and heuristics see Graham; Knuth; Patashnik: Concrete Mathematics. If you're seeking algorithms see Carsten Schneider's thesis Symbolic summation in Difference Fields, 2001, and Petkovsek; Wilf; Zeilberger: A = B. –  Bill Dubuque Apr 6 '11 at 4:58

1 Answer 1

Well, you can consider three ways, for one, use the inductive method; for two, try to write down what you want to sum, and then multiply it by 2 and subtract them; for three, try to write it as a recurrence relation, and compute it as usual. In each case, just remember that x-1 is a divisor of $x^n -1$ when n is a strictly positive natural number.

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