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{
A Note on an American Mathematical Monthly  Note
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\rm
In the Feb. 2011 issue of the {\it American Mathematical Monthly}, pp. 175-177, there is
a probabilistic proof (by G. Chang and C. Xu) of the identity $\sum_{i=0}^{n} {{2i} \choose {i}}{{2n-2i} \choose {n-i}}=4^n$,
and of a generalization. 
This, and the generalization, follow 
by extracting the coefficient of $x^n$ in $((1-4x)^{-1/2})^2=(1-4x)^{-1}$ and $((1-4x)^{-1/2})^m=(1-4x)^{-m/2}$ 
respectively.
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Doron Zeilberger, {\tt http://www.math.rutgers.edu/\Tilde zeilberg/}, Feb. 11, 2011.
({\eighttt Exclusively published in {\tt http://www.math.rutgers.edu/\Tilde zeilberg/pj.html} (Personal J. of S.B. Ekhad and D. Zeilberger)} .)

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