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I found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.

Prove for odd $n$ that $$\sum_{k=1}^{(n+1)/2}\prod_{j=0}^{k-1} \left(\frac{n-2j-1}{n-2j}\right)=\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$

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    $\begingroup$ As soon as you can define $\sum_{k=1}^n$ without recursion $\endgroup$
    – Hagen von Eitzen
    Commented Sep 29, 2017 at 6:11
  • $\begingroup$ @HagenvonEitzen I changed the limit. Was that the problem? $\endgroup$
    – videlity
    Commented Sep 29, 2017 at 6:15

1 Answer 1

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We need to prove that $$\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$ or $$1+\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n}{3}$$ or $$\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-3}{3}$$ or $$1+\frac{n-5}{n-4} +\frac{n-5}{n-4}\frac{n-7}{n-6} ... = \frac{n-2}{3}$$ or $$.$$ $$.$$ $$.$$ $$1+\frac{2}{3}=\frac{5}{3}.$$ Done!

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  • $\begingroup$ Plus 1 for the nice presentation. But I don't think you've avoided induction. $\endgroup$
    – B. Goddard
    Commented Sep 29, 2017 at 12:19
  • $\begingroup$ @B. Goddard Yes you are right of course. In each problem, where there are "...", we actually use an induction. I think it means to prove our identity without cases: 1. base; 2. assuming; 3. proof. If we want to use the idea of the telescopic sum then we also need to use induction, but I think it's just delirium to make it with previous cases 1;2 and3. $\endgroup$
    – Michael Rozenberg
    Commented Sep 29, 2017 at 12:31

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