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 $p+1$? – May 17 '13 at 23:03
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thanks alot Peter Tamaroff – mnsh May 17 '13 at 23:04
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is there any one help me and lead me to full answer? – mnsh May 17 '13 at 23:11
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1http://planetmath.org/proofoffaulhabersformula – anon May 18 '13 at 02:50
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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.
ADD This is Faulhaber's Formula.
Pedro
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1@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 May 17 '13 at 23:21
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This should be a quick application of Euler MacLaurin summation formula.
See my previous answer here: https://math.stackexchange.com/a/18989/1102
(Note: I haven't really verified if it is true).
