Solving Finite Series: \sum_{l=0}^{k-1} (r+l)^j (r+l-k)^i

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The discussion focuses on the finite series \(\sum_{l=0}^{k-1} (r+l)^j (r+l-k)^i\), where \(r\), \(k\), \(i\), and \(j\) are positive integers with \(i \leq j\). The main objective is to express this series as a polynomial in \(k\) and determine the coefficients as \(i\) and \(j\) vary. The user suggests that terms with even exponents may vanish, but a counterexample is provided showing that for \(i=2\) and \(j=3\), the term \(k^6\) appears, indicating that simplification is complex. Bernoulli polynomials are proposed as a potential avenue for expressing the series.

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ekkilop
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Hi!
I've encountered the series below:

\sum_{l=0}^{k-1} (r+l)^j (r+l-k)^i

where r, k, i, j are positive integers and i \leq j.
I am interested in expressing this series as a polynomial in k - or rather - finding the coefficients of that polynomial as i,j changes. I have reasons to believe that all terms with even exponent will vanish though I cannot readily see it from the expression above.

I have made some feeble attempts at expressing this in terms of well known functions. The most promising seems to be in terms of Bernoulli polynomials due to the large number of catalogued identities for these functions, though I haven't found a suitable one;

\frac{1}{(i+1)(j+1)} \sum_{l=0}^{k-1} [B_{j+1}(r+l+1) - B_{j+1}(r+l)][B_{i+1}(r+l-k+1) - B_{i+1}(r+l-k)]

Anyway, I cannot see how to proceed from here.
Any ideas or insights would be greatly appreciated.

Thank you!
 
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ekkilop said:
I have reasons to believe that all terms with even exponent will vanish though I cannot readily see it from the expression above.
I found a counterexample: i=2, j=3, r does not matter. You get k^6 (and maybe k^2, depends on r) as term.
WolframAlpha query

I have no idea how to simplify the sum.
 

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