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Prove algebraically or otherwise:

$$\sum \limits_{r=0}^n {2r \choose r} {2n-2r \choose n-r} = 4^n $$

where ${n \choose r}$ denotes the usual binomial coefficient. I think there is a combinatorial proof using path counting arguments, but I haven't been able to find it.

MathManiac
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1 Answers1

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Start with

$$(1+kx)^\alpha=1+\sum\limits_{j=1}^\infty{\alpha(\alpha-1)\cdots(\alpha-j+1)\over j!}k^jx^j$$

With $\alpha=-{1\over 2}$ and $k=-4$ the general term of the above series is

$${2k\choose k}={(k+1)\cdots 2k\over k!}$$

$\sum \limits_{r=0}^n {2r \choose r} {2n-2r \choose n-r}$ is just the coefficient of $x^n$ in the series Cauchy product

$$\left(\sum\limits_{k=0}^\infty{2k\choose k}x^k\right)^2$$

Now keep in mind that

$$\sum_{k=0}^\infty{2k\choose k}x^k={1\over\sqrt{1-4x}}$$

So our sum is the coefficient of $x^n$ in the series expansion of

$${1\over 1-4x}=\sum\limits_{k=0}^\infty 4^kx^k$$

so it is $4^n$ as requested.

marwalix
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  • How do you prove that $1/\sqrt{1-4x}$ is the generating function of $\left(\binom{2k}{k}\right)_{k=0}^\infty$? – Falko Sep 04 '15 at 10:02
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    I started with the development of $(1-kx)^\alpha$ and did some matching. I started my answer with ${2k\choose k}={(1+k)\cdots\over k!}$ for that reason. I am in the car (not driving) with my smartphone. I will edit as soon as I am at my desk – marwalix Sep 04 '15 at 10:16