# 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 $$p > 1$$?

• Is x intended to be an integer? Jun 7, 2012 at 14:47
• as far as i know there is no such a general formula to calculate $\sum_{k=1}^{n}k^x$ even for $x$ is an integer; Jun 7, 2012 at 14:48
• If $x$ is an integer, then Faulhaber's formula is what you're looking for. I believe this question has been asked before, but I can't find the duplicate. Jun 7, 2012 at 14:50
• Aug 11, 2018 at 13:00

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