Calculating a Finite Series: Finding Symmetry and Inductive Formulas

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The discussion focuses on calculating the finite sum \(\sum_{i=1}^{n-1}i^{\alpha}(n-i)^{\beta}\) for natural numbers \(\alpha\), \(\beta\), and \(n\). The user, Latrace, has attempted to derive an inductive formula by evaluating the sum for small values of \(n\) but found no significant results. The symmetry of the series is noted, as it can also be expressed as \(\sum_{i=1}^{n-1}i^{\beta}(n-i)^{\alpha}\). Ultimately, the user seeks to understand the behavior of the sum as \(n\) becomes large, approximating it to \((n-1)^{\beta} + (n-1)^{\alpha}\). The discussion highlights the challenges in finding a general formula for the series.
Latrace
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Hello,

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

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

Thanks in advance,
Latrace
 
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What did you already try to solve this problem?
 
(n \geq 2, of course) I tried to find an inductive formula by setting n = 2, n = 3 and n = 4, but don't find anything interesting. Of course we already knew that the thing is symmetric, symbolically it is also \displaystyle\sum_{i=1}^{n-1}i^{\beta}(n-i)^{\alpha}, 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 n, but that's \sim (n-1)^{\beta} + (n-1)^{\alpha}. I came across this when I wanted to calculate \displaystyle\int_{0}^{1}x^m \mathrm{d}x for m \geq 1 explicitally using the Riemann sum.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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