I know from my calculator the answer is $\sum_{i = 1}^n k^n$ = $\frac{k^{n+1}-k}{k - 1}$. I'd just like help understanding why.
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Just check that $k^n -1 = (k-1)(\sum_{i=0}^{k=n-1} k^i)$ – Andrei Kulunchakov Jul 03 '15 at 17:32
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Your index is $i$ so $\sum_{i=1}^nk^n=nk^n$. You probably mean $\sum_{k=1}^nk^n$. Or did you mean $\sum_{i=1}^Nk^i$? – Mark Viola Jul 03 '15 at 17:44
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Because $$(1-k)(1+k^2+k^3+\dots+k^n)=(1+k^2+k^3+\dots+k^n)-(k+k^2+k^3+\dots+k^{n+1})=1-k^{n+1}$$
FrodCube
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