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Let $l,p,b$ be positive integers. The question is to compute a following sum: \begin{equation} (1) {\mathcal S}_{p,l}(b) := \sum\limits_{i=1}^b i^p \left(\begin{array}{r} i \\ l \end{array}\right) \end{equation} For $p=0$ the the result is trivial and reads ${\mathcal S}_{0,l}(b) = \left(\begin{array}{r} b+1 \\ l+1 \end{array} \right)$. Now, the idea is to decompose the factor $i^p$ into a linear combination of Pochhammer symbols as follows: \begin{equation} (2) i^p = (i+p)_{(p)} - \left(\begin{array}{c} p+1 \\ 2 \end{array} \right) (i+p-1)_{(p-1)} + \left(\begin{array}{c} p+1 \\ 3 \end{array} \right) \frac{3 p-2}{4} (i+p-2)_{(p-2)} -\left(\begin{array}{c} p+1 \\ 4 \end{array}\right) \frac{(p-1)(p-2)}{2} (i+p-3)_{(p-3)} + \dots \end{equation} Then we insert the above into the definition of ${\mathcal S}_{p,l}(b)$, combine the Pochhammer symbols with the binomial factor that involves $i$ and perform the sum using the result for ${\mathcal S}_{0,l}(b)$. This gives the following: \begin{eqnarray} (3) {\mathcal S}_{p,l}(b) = \\ &&\frac{(l+p)!}{l!} \left(\begin{array}{c} b+p+1 \\ l+p+1 \end{array}\right) + \frac{1}{l!} \sum\limits_{q=2}^p (-1)^{q-1} (l+p-q+1)! \left(\begin{array}{c} p+1 \\ q \end{array}\right) \cdot {\mathcal F}(q) \cdot \left(\begin{array}{c} b+p-q+2 \\ l+p-q+2 \end{array}\right) \end{eqnarray} where the factors ${\mathcal F}(q)$ read: \begin{equation} \left({\mathcal F}(2),{\mathcal F}(3),\dots\right) = \left(\frac{3 p-2}{4},\frac{(p-1)(p-2)}{2},\dots\right) \end{equation} Equation (3) can be seen as a large-$b$ expansion of our sum. Now, the question is can we prove (3) using some other method and can we find a closed form expression for the factors ${\mathcal F}(q)$?

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