Does anyone know how to show the following combinatorial equality, $\sum_{i=0}^{n}\left(n-i\right)^{2}\binom{2n}{i}=n\cdot4^{n-1}$?
By the way, this is not a homework problem, otherwise one would be able to search the answer.
Thanks.
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4$\begingroup$ As Robin Chapman mentions, your sum is half of the full sum $\sum_{i=0}^{2n}(n-i)^2\dbinom{2n}i$. Note that $\sum_{i=0}^{2n}\dbinom{2n}ix^i=(1+x)^{2n}$ and if you aplly the differential operator $x\dfrac{d}{dx}$, you also have the closed formulas for $\sum_{i=0}^{2n}i^l\dbinom{2n}ix^i$ with $l=1,2$. A linear combination of the three will allow you to compute your expression. It's a homework problem! $\endgroup$– Wadim ZudilinCommented May 23, 2010 at 9:50
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3 Answers
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It's half the sum of the same thing from $0$ to $2n$, which in turn is easily related to the variance of the number of heads in a sequence of $2n$ tosses of a fair coin.
answered May 23, 2010 at 9:37
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1$\begingroup$ Robin, I spent time following your combinatorial hint. If the author claims that his problem isn't a homework, your combinatorial solution is definitely best possible. +1 $\endgroup$ Commented May 23, 2010 at 11:59
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Try the Wilf-Zeilberger method and its friends. This automatically proves many such (hypergeometric) identities. See the book A = B
answered May 23, 2010 at 9:36
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I suggest you take a look on hypergeometric series.
answered May 23, 2010 at 9:38