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Prove that prove $\dbinom{4n}{2n} = \dfrac{1\cdot3\cdot5\cdots(4n-1)}{(1\cdot3\cdot5\cdots(2n-1))^2} \dbinom{2n}{n}$ using mathematical induction. I have looked all over the internet, been able to prove a similar problem, but this one has me stumped. Quick help would be appreciated! Thanks!

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Hint: $$\binom{4(n+1)}{2(n+1)} =\frac{(4n+4)(4n+3)(4n+2)(4n+1)}{((2n+2)(2n+1))^2}\binom{4n}{2n}$$

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  • $\begingroup$ Okay, does the right hand side come down to $\frac{4(4n+3)(4n+1)}{(2n+2)(2n+1)} {4n \choose 2n}$ ? $\endgroup$ – Ashley Apr 14 '15 at 23:40
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Hints:

  • $\dfrac{1\times 3\times 5\times \cdots \times (4n-1)}{(1\times 3\times 5\cdots \times (2n-1))^{2}} $ $= \dfrac{1\times 2\times 3\times 4\times 5\times \cdots \times (4n-1)\times 4n}{(1\times 2\times 3\times 4\times 5\times \cdots \times (2n-1)\times 2n)^{2}} \times \dfrac{(2\times 4\times \cdots \times 2n)^{2}}{2\times 4\times \cdots \times 4n}$

  • $\dfrac{1\times 2\times 3\times 4\times 5\times \cdots \times (4n-1)\times 4n}{(1\times 2\times 3\times 4\times 5\times \cdots \times (2n-1)\times 2n)^{2}} = \dfrac{(4n)!}{(2n!)^2}$

  • $\dfrac{(2\times 4\times \cdots \times 2n)^{2}}{2\times 4\times \cdots \times 4n}=\dfrac{(2^n\times 1\times 2\times \cdots \times n)^{2}}{2^{2n}\times 1\times 2\times \cdots \times 2n}=\dfrac{(n!)^2}{(2n)!}$

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