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I dont know how to proceed with solving $$\sum_{i=1}^{n}i^{k}(n+1-i).$$ Please give advise.

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@RossMillikan In the summand, the power is a constant... –  David Mitra Apr 12 '12 at 16:07
@DavidMitra: You are correct. It is not a duplicate. –  Ross Millikan Apr 12 '12 at 16:46
First, find two generating functions for $n^k$ and $1+n$. Then product of them. Next, get the coefficient of $x^n$. You need the eulerian number for the generating function of the sequence $n^k$. –  hkju Apr 13 '12 at 15:54

4 Answers 4

You can factor out the $(n+1)$ to give $(n+1)\sum_{i=1}^n i^k-\sum_{i=1}^n i^{k+1}$ For positive integral $k$ you can use Faulhaber's formulas. What kind of $k$ are you considering?

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is same as


which looks like some combination of Eulerian number.

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$H_n^{(k)}=\sum_{i=1}^n i^{-k}$ is by definition the $n$-th harmonic number of order $k$. Thus, $$ \sum_{i=1}^n i^{k} (n+1-i) = (n+1) H_n^{(-k)}- H_n^{(-k-1)}. $$ I don't think it can be simplified further, at least considerably and for any $n$ and $k$. Is this what you meant by "solving"?

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Ross: observe that $k$ is not the summation index. –  andreimf Apr 12 '12 at 16:03

This is a prefix sum of natural numbers. The solution to this is


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