<s>[INST] Finite Sum of Power? Can someone tell me how to get a closed form for
$$\sum_{k=1}^n k^p$$
For $p = 1$, it's just the classic $\frac{n(n+1)}2$. 
What is it for $p > 1$?
 [/INST] By Binomial Series
$(n+1)^x=1 + {x \choose 1}\sum_{k=1}^n{k^{x-1}} + {x \choose 2}\sum_{k=1}^n{k^{x-2}}+{x \choose 3}\sum_{k=1}^n{{k^{x-3}}}
...+{x \choose x-1}\sum_{k=1}^n{{k^{x-x+1}}}+{x \choose x}\sum_{k=1}^n{{k^{x-x}}}$
which becomes
$(n+1)^x = 1 + {x \choose 1}\sum_{k=1}^n{k^{x-1}} + {x \choose 2}\sum_{k=1}^n{k^{x-2}}+{x \choose 3}\sum_{k=1}^n{{k^{x-3}}}
....+{x \choose x-1}\sum_{k=1}^n{{k}}+{x \choose x}\sum_{k=1}^n{{1}}$
for example consider $x=3$
$(n+1)^3 = 1 + {3 \choose 1}\sum_{k=1}^n{k^{2}} + {3 \choose 2}\sum_{k=1}^n{k^{1}}+{3 \choose 3}\sum_{k=1}^n{{1}}$
$(n+1)^3 = 1 + 3\sum_{k=1}^n{k^{2}} + 3*\frac{n*(n+1)}{2}+n$
which gives
$\sum_{k=1}^n{k^2} =(1/6)*n*(n+1)*(2n+1)$
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