I am asked to prove that [tex] \sum ^n _{k=0} (C^n_k)^2 = C^{2 n}_n[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Where [tex] C^n_k [/tex] signifies "n choose k"

I am told the hint to use the binomial theorem and to calculate the coefficient of x^n in the product [tex](1+x)^n (1+x)^n = (1+x)^{2n}[/tex]

the Binomial theorem is given by [tex] (x+y)^n = \sum_{k=0} ^n C_k^n x^{n-k} y^k [/tex]

Following the hint, we see that:

[tex](1+x)^n = \sum^n _{k=0} C^n _k x^{n-k} [/tex]

[tex](1+x)^{2n} = \sum^{2n} _{k=0} C^{2n} _k x^{2n-k} [/tex]

I believe from here, I would like to show that:

[tex]\left( \sum^n _{k=0} C^n _k x^{n-k} \right) \left( \sum^n _{k=0} C^n _k x^{n-k} \right) = \sum^{2n} _{k=0} C^{2n} _k x^{2n-k}[/tex]

And ultimately, I believe the goal is to get [tex] \sum ^n _{k=0} (C^n_k)^2 = C^{2 n}_n[/tex] from the above relation but I'm not sure where to go from here.

Any ideas?

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# Homework Help: Proof using binomial theorem

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