# Identity with Bernoulli numbers: $\sum\limits_{k=1}^{n}k^p=\frac{1}{p+1}\sum\limits_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j}$

How I can prove that $$\sum_{k=1}^{n}k^p=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j},$$ where $B_j$ is the $j$th Bernoulli number?

I hope to find the answer. Thanks for help.

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In some ways, the coefficients that appear in this are the definition of Bernoulli numbers... So are you interested in knowing why $\sum_{k=1}^n k^p$ is a polynomial in $n$ of degree $p+1$? –  user17762 May 17 '13 at 23:03
thanks alot Peter Tamaroff –  mhd.math May 17 '13 at 23:04
is there any one help me and lead me to full answer? –  mhd.math May 17 '13 at 23:11

This is a consequence of the use of the Bernoulli polynomials. We can define them by $$B_0(x)=1$$ $$B'_{n}(x)=nB_{n-1}(x)$$and $$\int_0^1 B(x)dx=0$$

They have the particular property that $$B_{p+1}(x+1)-B_{p+1}(x)=(p+1)x^{p}$$ which is why we use them to evaluate such a sum, also $B_{n}(0)=B_{n}$, the $n$-th Bernoulli number.

You can find any $B_n(x)$ with the above, and thus evaluate any sum $$\sum_{k=1}^nk^p$$

Give it a try! Set $x=k$ and sum away, to obtain that $$\sum\limits_{k = 1}^n {{k^p} = } \frac{{{B_{p + 1}}\left( {n + 1} \right) - {B_{p + 1}}\left( 0 \right)}}{{p + 1}}$$

In fact, to get your sum to look like in your question you will have to show that $$B_n(x)=\sum_{j=0}^n \binom{n}{j}B_jx^{n-j}$$ where $B_j:=B_j(0)$. Induction should be enough.

@mohammad.mnsh I haven't got the time now, try it yourself. Maybe some other day I can add something, but it is nice to work things out on your own. Try to obtain some of the polynomials for example, and see what happens in each case $p=1,2,3,4$, say, to get a taste of what is going on. There are surely many, many pages where you can find proofs of this, just Google "Faulhaber's formula." –  Pedro Tamaroff May 17 '13 at 23:21