7

Show that

$$\sum_{k=0}^n \left[ \frac{n-2k}{n} {n\choose k}\right]^2=\frac{2}{n}{2n-2 \choose n-1}.$$

Paul S. Quail
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    You mention contest-math. From what contest did this come? What have you tried? – robjohn Jul 13 '13 at 22:08
  • A translation in terms of random walks simplifies the computations, partly. – Did Jul 13 '13 at 22:15
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    I always considered the contest-math tag to be for problems in the contest style, not just those from actual contests. Perhaps this is the case here? – Potato Jul 13 '13 at 22:19
  • @Did We can do that with the Chu-Vandermonde identity (as shown by robjohn's deleted answer). But a translation in terms of random walks would make the question more interesting for me. What would that be? – Julien Jul 13 '13 at 22:32
  • @Potato: I mainly wanted to verify that this was not from an ongoing contest. Once I know that, I will undelete my answer. It would be nice to see what Did was suggesting, too. – robjohn Jul 13 '13 at 23:27
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    @robjohn Is the default presumption really that all tricky-looking problems are from ongoing contests? In any case, I can verify that this one isn't. It's from Problem Solving Through Problems by Larson. – Potato Jul 14 '13 at 00:01
  • @Potato: that it appears in a book doesn't mean that it isn't also in an ongoing contest, but this doesn't seem like a contest problem. However, in addition, this is also a PSQ. I'll wait for a bit more and see if the OP has anything to add. – robjohn Jul 14 '13 at 00:47
  • @Potato: Thanks for the source of the problem ;-) – robjohn Jul 14 '13 at 03:06
  • @robjohn ;-) $\textbf{}$ – Potato Jul 14 '13 at 06:15
  • @julien The LHS times $n^22^{-2n}$ is $E[S_n^2;S_{2n}=0]$, where $(S_k)$ is a standard $\pm1$ random walk. Expanding $S_n^2$, this is $nP[S_{2n}=0]$ (known) plus $n(n-1)\alpha$ with $\alpha=E[X_1X_2;S_{2n}=0]$, where $(X_k)$ are the increments of the random walk. Since $X_1X_2=\pm1$, $\alpha$ is $\frac12P[S_{2n-2}=2]$ (known) minus $\frac12P[S_{2n-2}=0]$ (known). Reducing the three binomial coefficients involved to ${2n-2\choose n-1}$ yields the result, but this last part of the proof, while easy, is not very illuminating nor very random-walk-oriented, which is the reason why I wrote "partly". – Did Jul 14 '13 at 08:02
  • .../... However, somebody cleverer than me might find a way to compute $E[S_n^2;S_{2n}=0]$ using purely probabilistic arguments till the end... – Did Jul 14 '13 at 08:03
  • @Did this random walk thing is pretty new to me, though I do have sort of a dim idea. Any resources for a newb? :) – Soham Chowdhury Jul 14 '13 at 09:59
  • @SohamChowdhury A short one and a (much) longer one. – Did Jul 14 '13 at 10:57
  • @Did Ok, thanks a lot. – Julien Jul 14 '13 at 13:33
  • @julien You are welcome. I now asked this as a separate question. – Did Jul 14 '13 at 13:55
  • @Did the second link 404s. – Soham Chowdhury Jul 15 '13 at 15:35
  • @SohamChowdhury Sorry, try this one. – Did Jul 15 '13 at 23:56

1 Answers1

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$$ \begin{align} \sum_{k=0}^n\left[\frac{n-2k}{n}\binom{n}{k}\right]^2 &=\sum_{k=0}^n\left[\binom{n}{k}-2\binom{n-1}{k-1}\right]^2\\ &=\sum_{k=0}^n\left[\binom{n-1}{k}-\binom{n-1}{k-1}\right]^2\\ &=\sum_{k=0}^n\binom{n-1}{k}^2+\binom{n-1}{k-1}^2-2\binom{n-1}{k}\binom{n-1}{k-1}\\ &=2\binom{2n-2}{n-1}-2\sum_{k=0}^n\binom{n-1}{k}\binom{n-1}{n-k}\\ &=2\binom{2n-2}{n-1}-2\binom{2n-2}{n}\\ &=2\binom{2n-2}{n-1}-2\frac{n-1}{n}\binom{2n-2}{n-1}\\ &=\frac2n\binom{2n-2}{n-1} \end{align} $$

robjohn
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