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Define Bernoulli polynomials as: $P_0(x)=1$, $P'_n(x)=nP_{n-1}(x)$, $\int_0^1P_n(x)=0$ if $n\geq1$

Need to prove that for $n\geq2$ we have

$$\sum_{r=1}^{k-1} r^n= \frac{P_{n+1}(k)-P_{n+1}(0)}{n+1}$$

I have already proved that:

$P_n(1)=P_n(0)$ if $n\geq2$ and $P_n(x+1)-P_n(x)=nx^{n-1}$ if $n\geq1$

I tried using induction, but got stuck making the inductive step. Could anybody provide any hints? Thanks!

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up vote 2 down vote accepted

If $$P_n(x+1)-P_n(x)=nx^{n-1}$$ then $$\sum_{x=0}^{k-1} [P_n(x+1)-P_n(x)]=\sum_{x=0}^{k-1} nx^{n-1}$$ but the LHS telescopes to $P_n(k)-P_n(0)$.

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yes, that is clear enough. thank you! –  Sarunas Nov 23 '12 at 22:08
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