General formula for summing numbers to odd powers

I found for $l$ odd powers

$\sum_{k=1}^n k^l= \frac{1}{l+1}n^{l+1}+\frac{1}{2}n^{l} +\frac{9+l}{12}n^{l-1} +-\zeta(-3) \binom{2(n-1)+3}{3}n^3 + S(l)$

Where

$S(l) = \sum_{k=1}^{l-2} -\zeta(-k)\binom{2n+k}{k}$

Seems to work. Although the first terms are different, maybe I can express it in all like the last one.

The sum can also be expressed in the following form:

$S(l) = \sum_{k=5}^{l-2}\prod_{k = 1}^l a_k (n+k)(2n+2k+1)$

And the whole sum can be expressed in this form

$\sum_{k=1}^n k^l= \sum_{k=1}^n b_k (n^2+n)^k$

But I don’t know $a_k, b_k$ yet, maybe I can find those coefficients with the zeta function

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