I'm looking for the generating function of the sum $\sum_{i=1}^{n-k} {n-1-i\choose k-1}x^i$. One can compute this using the Euler-MacLauren formula but the remainder term is a little messy. Is there a quick or easy way to compute this completely? I ask because a closed form of this would imply the closed form of the $(n^m,1)$ Pascal triangles, and the basic formula for the $(a_n,b_n)$ Pascal triangles doesn't give the full closed form.
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asked Jun 16, 2021 at 21:27
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2$\begingroup$ A closed formula does not seem feasible here. $\endgroup$– T. AmdeberhanCommented Jun 16, 2021 at 21:42
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$\begingroup$ You have already asked this in Closed form of $ \sum_{i=1}^{n-k} {n-1-i\choose k-1}i^a + \sum_{i=1}^k {n-1-i\choose n-1-k}$ $\endgroup$– Max AlekseyevCommented Jun 16, 2021 at 21:54
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$\begingroup$ This question is different in structure. $\endgroup$– Benjamin L. WarrenCommented Jun 16, 2021 at 21:55
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$\begingroup$ Then you need elaborate what is different and/or why you are not happy with answers given there. $\endgroup$– Max AlekseyevCommented Jun 16, 2021 at 21:58
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$\begingroup$ In that question I was looking for ${n-2\choose k-1}1^a + {n-3\choose k-1}2^a...+{k-1\choose k-1}(n-k)^a$ but here I'm looking for ${n-2\choose k-1}x + {n-3\choose k-1}x^2...+{k-1\choose k-1}x^{n-k}$. $\endgroup$– Benjamin L. WarrenCommented Jun 16, 2021 at 22:01
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1 Answer
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It depends on what your notion of "closed" is. A quick calculation in Mathematica shows that your expression equals
$x{n-2\choose k-1}\cdot {}_2 F_1(1,1+k-n;2-n;x)$,
where ${}_2 F_1$ denotes the Gaussian hypergeometric function. These are well-understood in many contexts.
answered Jun 18, 2021 at 6:41