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I asked this questions on the https://math.stackexchange.com/questions/1016713/ But I don't get answer. I try to prove the following hypothesis $$\sum_{i=0}^{min\{k, n-1\}}(-1)^i { n+i-1 \choose i}{{n+k-2} \choose k-i} {n+2k-i-1 \choose 2k}= \sum_{i=0}^{\min\{k, n-3\}}{{n-3} \choose i}{{n-2} \choose i} {{2n+k-i-4} \choose {k-i}}\frac{1}{i+1}$$ Calculation in the Maple shows that $$\sum_{i=0}^{min\{k, n-1\}}(-1)^i { n+i-1 \choose i}{{n+k-2} \choose k-i} {n+2k-i-1 \choose 2k}= \sum_{i=0}^{\min\{k, n-3\}}{{n-3} \choose i}{{n-2} \choose i} {{2n+k-i-4} \choose {k-i}}\frac{1}{i+1}=\sum_{i=0}^{k}(-1)^i { n+i-1 \choose i}{{n+k-2} \choose k-i} {n+2k-i-1 \choose 2k}= \sum_{i=0}^{k}{{n-3} \choose i}{{n-2} \choose i} {{2n+k-i-4} \choose {k-i}}\frac{1}{i+1}= {\frac {{n+k-1\choose k}{n-2+k\choose k}}{k+1}}$$ Here are some attempts.- $$ \sum_{i=0}^{min\{k, n-1\}}(-1)^i { n+i-1 \choose i}{{n+k-2} \choose k-i} {n+2k-i-1 \choose 2k}=\\= \sum_{i=0}^{min\{k, n-1\}} { -n \choose i}{{n+k-2} \choose k-i} {n+2k-i-1 \choose 2k}=\\= \sum_{i=0}^{min\{k, n-1\}} \frac{(-1)^i(n+i-1)(n+k-2)!}{i!(k-i)!(n-1)!} {n+2k-i-1 \choose 2k} =\\=\sum_{i=0}^{min\{k, n-1\}} \frac{(-1)^i(n+i-1)(n+k-1)!k!}{i!(k-i)!k!(n-1)!(n+k-1)} {n+2k-i-1 \choose 2k}= \\ = {n+k-1 \choose k}\sum_{i=0}^{min\{k, n-1\}} \frac{(-1)^i(n+i-1)}{(n+k-1)} {k \choose i} {n+2k-i-1 \choose 2k}.$$ This is my attempt so far. I tried use Mathematical induction method (about k) and Chu's Theorem. The second way was using formula ${r+k+1 \choose k}=\sum_{i=0}^k {r+i \choose i}.$ I cannot prove it. Help me, please.

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  • $\begingroup$ Dividing it up in a few cases, each of them should be provable (via computer) with Zeilbergers algorithm. $\endgroup$ Nov 12, 2014 at 11:14
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    $\begingroup$ if I understand the question correctly, Maple has already proven this identity, and even given a closed form expression for the sum, presumably using Zeilberger. $\endgroup$ Nov 12, 2014 at 14:52
  • $\begingroup$ @Per Alexandersson I can not use Zeilbergers algorithm in Maple. please tell me what files need I download? $\endgroup$
    – Liza
    Nov 17, 2014 at 13:14
  • $\begingroup$ Zeilbergers algorithm is build-in in Maple. GIYF. maplesoft.com/support/help/Maple/view.aspx?path=SumTools/… $\endgroup$ Nov 17, 2014 at 13:56

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