I dont know how to proceed with solving $\sum_{i=1}^{n}i^{k}(n+1-i).$ Please give advise.
Computing $\sum_{i=1}^{n}i^{k}(n+1-i)$
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sequences-and-series
summation
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0First, 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 $n^k$. – 2012-04-13
4 Answers
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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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$H_n^{(k)}=\sum_{i=1}^n i^{-k}$ 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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0Ross: observe that $k$ is not the summation index. – 2012-04-12
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This is a prefix sum of natural numbers. The solution to this is
$\binom{n+k+1}{k+2}$