# Formula for $\sum_{k=1}^{n}{k^p}$ where p is a positive integer [duplicate]

Possible Duplicate:
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$

Any hints that can take me from here or am I completely lost.

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a-1}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{a1<a2<a3...<an}a1)$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a-1}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{i=1}^{n}(n+1-i)^{a-1}(i)))$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{i=1}^{n}(n+1-i)^{a-1}(i)))$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(a!S(n,a))(\sum_{i=1}^{n}(i)^{a-1}(n+1-i)))$

Where S(n,a) is a stirling number of second kind.

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## marked as duplicate by Aryabhata, Did, anon, Eric Naslund, t.b.Apr 12 '12 at 11:21

This question was marked as an exact duplicate of an existing question.

I assume $p$ is a positive integer. There is a moderately complicated General Formula for the sum, involving the Bernoulli numbers. The term "power series" is a technical term that refers to something else. – André Nicolas Apr 12 '12 at 5:39
Yes p is a positive integer. Thanks for correcting me. – bspk Apr 12 '12 at 5:44
If you like you look at go.helms-net.de/math/potenzsummen/potenzsummen_1.htm which is a very old and completely elementary treatize of mine, of when I explored this myself the first time. Unfortunately in german, but I think the formulae and expressions are clear enough to hint you to a fruitful direction (for instance introducing how the Eulerian numbers of hkju's answer come into play) – Gottfried Helms Apr 12 '12 at 9:50

$$\sum_{k=1}^n k^p = \sum_{k=0}^{p-1} T(p,k)\binom{n+1+k}{p+1}. \mbox{ (hence, its degree is p+1)}$$, where $T(p,k)$ is the Eulerian number (cf.OEIS A008292). For example, $p=2$:$$\sum_{k=1}^n k^2 = \sum_{k=0}^{1} T(2,k)\binom{n+1+k}{3}=(1)\binom{n+1}{3}+(1)\binom{n+2}{3}=\frac{n(n+1)(2n+1)}{6}$$ $p=3$:$$\sum_{k=1}^n k^3 = \sum_{k=0}^{2} T(3,k)\binom{n+1+k}{4}=(1)\binom{n+1}{4}+(4)\binom{n+2}{4}+(1)\binom{n+3}{4}=\frac{n^2 (n+1)^2}{4}$$