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 $p > 1$?


The general formula is fairly complicated:

$$\sum_{k=1}^n{k^x} = {1\over x+1}\sum_{k=0}^x{{x+1\choose k}B_k n^{x+1-k}}$$

Where $B_k$ are the Bernoulli numbers, discovered around the same time, but independently, by Johann Bernoulli and Seki Takakazu.

This is commonly called "Faulhaber's formula", but that is a misattribution. The Donald Knuth paper "Johann Faulhaber and sums of powers" explains the history in some detail, including what Faulhaber did and did not know. In particular, he did not know the general formula above, although he did work on many special cases.

The first few special cases are:

$$\begin{array}{ll} \sum_{k=1}^n 1 & = n \\ \sum_{k=1}^n k & = \frac12{(n^2+n)} \\ \sum_{k=1}^n k^2 & = \frac16{(2n^3+3n^2+n)} \\ \sum_{k=1}^n k^3 & = \frac14{(n^4+2n^3+n^2)} \\ \sum_{k=1}^n k^4 & = \frac1{30}{(6n^5+15n^4+10n^3-n)} \\ \sum_{k=1}^n k^5 & = \frac1{12}{(2n^6+6n^5+5n^4-n^2)} \\ \end{array}$$


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 $\sum_{k=1}^n{k^2} =(1/6)*n*(n+1)*(2n+1)$

  • $\begingroup$ I think the questioner was looking not just for when $x=2, $ But for any $x\gt 1$ $\endgroup$ – amWhy Dec 5 '16 at 23:42
  • $\begingroup$ X=2 was an example. For any X you need to know results of x-1,x-2,x-3... $\endgroup$ – dshri Dec 5 '16 at 23:47
  • $\begingroup$ Very good. Your answer is safe! $\endgroup$ – amWhy Dec 5 '16 at 23:49

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