Calculating a Finite Series: Finding Symmetry and Inductive Formulas

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The discussion focuses on calculating the finite series \(\sum_{i=1}^{n-1}i^{\alpha}(n-i)^{\beta}\) for natural numbers \(\alpha\), \(\beta\), and \(n\). The user, Latrace, explores inductive formulas by evaluating the series for \(n = 2\), \(n = 3\), and \(n = 4\), noting its symmetry, as it can also be expressed as \(\sum_{i=1}^{n-1}i^{\beta}(n-i)^{\alpha}\). The key conclusion is that for large \(n\), the series behaves asymptotically like \((n-1)^{\beta} + (n-1)^{\alpha}\), which was derived while calculating the Riemann sum for \(\int_{0}^{1}x^m \mathrm{d}x\) for \(m \geq 1\).

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Latrace
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Hello,

I would love some help on calculating the following sum for [itex]\alpha, \beta \in \mathbb{N}[/itex] and [itex]n \in \mathbb{N} \backslash \{0\}[/itex]:

[itex]\displaystyle\sum_{i=1}^{n-1}i^{\alpha}(n-i)^{\beta}.[/itex]

Thanks in advance,
Latrace
 
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What did you already try to solve this problem?
 
([itex]n \geq 2[/itex], of course) I tried to find an inductive formula by setting [itex]n = 2, n = 3[/itex] and [itex]n = 4[/itex], but don't find anything interesting. Of course we already knew that the thing is symmetric, symbolically it is also [itex]\displaystyle\sum_{i=1}^{n-1}i^{\beta}(n-i)^{\alpha}[/itex], but that's about all I find when I try to find an inductive formula. I think now that this might be the easiest way to express the series.
What I eventually need is the behavior for large [itex]n[/itex], but that's [itex]\sim (n-1)^{\beta} + (n-1)^{\alpha}[/itex]. I came across this when I wanted to calculate [itex]\displaystyle\int_{0}^{1}x^m \mathrm{d}x[/itex] for [itex]m \geq 1[/itex] explicitally using the Riemann sum.
 

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