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$$ s_n(p)=\sum_{k=1}^n k^p $$

Show: For every $q \geq 1$ exist rational numbers $ a_{k,q} , 1 \leq k \leq q-1 $, such that

$$ s_n(q)= \frac 1 {q+1} n^{q+1}+ \frac 1 2 n^q + \sum_{k=1}^{q-1} a_{k,q}n^{q-k} $$

I am afraid I don't have a clue on how to prove that. Induction? Rearranging? Before that task the Pascal Identity was introduced:

$$ \sum_{p=0}^q \binom{q+1}p s_n(p)=(n+1)^{q+1} - 1 $$

But I can't really use that, because $s_n(p)$ refers to p and not to q, doesn't it?

Have you got any ideas on how to prove that?

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  • $\begingroup$ Hint: Put the Binomial Theorem expansion of $(k+1)^{p+1}$ into $\frac {1}{p+1}((k+1)^{p+1}-k^{p+1})$ and then sum over $k=1,...n.$ Then use induction on $p.$ $\endgroup$ – DanielWainfleet Apr 19 at 12:19
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Prove this is true for $s_n(0)$, which should be easy. Then, given $q$, write $$(q+1)s_n(q)=-\sum_{p=0}^{q-1} \binom{q+1}p s_n(p)+(n+1)^{q+1} - 1$$ There are a few more nontrivial steps to complete from here, but this at least tells you it is a polynomial in $n$.

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