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I'm having a bit of trouble with this proof and was wondering if I could get some assistance. I need to show that: $$ΦS(x) = \frac{1}{\sqrt{1-4x}}$$ All I have is: $$\sum\limits_{k=0}^n {2n\choose k}{2(n-k)\choose n-k} = 4^n$$

I know by definition that $$ΦS(x) = \sum\limits_{n\ge0}s_n\cdot x^n$$ the exponent is the weight and the coefficient is the number of elements with that weight. If I could get some help that would be great!

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up vote 2 down vote accepted

Your question isn’t very clear, but I take it that you’re trying to show that

$$\frac1{\sqrt{1-4x}}=\sum_{n\ge 0}\binom{2n}nx^n\;,\tag{1}$$

and you know the identity

$$\sum\limits_{k=0}^n {2n\choose k}{2(n-k)\choose n-k} = 4^n\;.\tag{2}$$

Let $$f(x)=\sum_{n\ge 0}\binom{2n}nx^n\;;$$ then

$$\begin{align*} \left(f(x)\right)^2&=\left(\sum_{n\ge 0}\binom{2n}nx^n\right)^2\\ &=\sum_{n\ge 0}\left(\sum_{k=0}^n\binom{2k}k\binom{2(n-k)}{k-k}\right)x^n\;, \end{align*}$$

where in the last step I’ve simply taken the Cauchy product. By $(2)$, therefore,

$$\left(f(x)\right)^2=\sum_{n\ge 0}4^nx^n=\sum_{n\ge 0}(4x)^n\;.$$

From here you should be able to derive $(1)$ without too much trouble.

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