# Computing $\sum_{i=1}^{n}i^{k}(n+1-i)$

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

$H_n^{(k)}=\sum_{i=1}^n i^{-k}$ 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

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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$$\sum_{i=1}^{n}i^{k}(n+1-i)$$

is same as

$$\sum_{i=1}^{n}i(n+1-i)^{k}$$

which looks like some combination of Eulerian number.

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

$$\binom{n+k+1}{k+2}$$

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