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$$\text{Prove That,}\;f(n) = \prod_{i=1}^{n}(4i - 2) = \frac{(2n)!}{n!}$$

This is a problem from Elementary Number theory.

My Work:
The statement is true for $i=1$. So, if it is true for $i=k$ it must be true for $ i = k+1$ . I am stuck here, any hint will be helpful.

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  • $\begingroup$ Pretty sure one of those $n$'s should be an $i$. Otherwise you have $(4n-2)^n$. Has no one caught this? $\endgroup$ – Mike Dec 28 '16 at 11:21
  • $\begingroup$ @Mike Then all the solutions are wrong also. Edited now. You are right (y) $\endgroup$ – Rezwan Arefin Dec 28 '16 at 11:22
  • $\begingroup$ Actually, I first noticed it in Rohan's answer. I was wondering why one $n$ got replaced with $n+1$ while another did not, then noticed something didn't look right. That still isn't correct by the way. The right side was fine as it was. $\endgroup$ – Mike Dec 28 '16 at 11:29
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To prove that it is true for $i =n+1$:

$$\prod_{i=1}^{n+1} (4i-2) = \prod_{i=1}^{n} (4i-2)(4(n+1)-2) = \frac{(2n)!}{n!} (4n+2) =\frac{(2n)!}{n!} 2(2n+1) =\frac{(2n)!}{n!}\frac{2n+2}{n+1}(2n +1) = \frac{(2n+2)!}{(n+1)!} = \frac{(2(n+1))!}{(n+1)!}$$ which was to be proved. Hope it helps.

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  • $\begingroup$ How $(2n+2)! = 2(n+1)!$ ? $\endgroup$ – Rezwan Arefin Dec 28 '16 at 10:49
  • $\begingroup$ @RezwanArefin It should be $(2n + 2)! = (2(n + 1))!$. $\endgroup$ – TheGeekGreek Dec 28 '16 at 10:50
  • $\begingroup$ @TheGeekGreek :P $\endgroup$ – Rezwan Arefin Dec 28 '16 at 10:51
  • $\begingroup$ The $n$'s should be $i$'s :P $\endgroup$ – Rezwan Arefin Dec 28 '16 at 11:23
  • $\begingroup$ @Rezwan Arefin Sorry for the mistake. In fact I followed on your steps. I will correct it now. $\endgroup$ – Rohan Dec 28 '16 at 11:27
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You can prove this without induction also,you just need to note that: $$\frac {f(n)}{f(n-1)}=4n-2$$ Hence $f(n)=(4n-2)f(n-1)$ Can you complete from here?

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Assume it is true for $n > 1$. Then we have by induction hypothesis $$\begin{align*}f(n + 1) &= \prod_{i = 1}^{n + 1} (4n - 2)\\ &= (4n + 2)\prod_{i = 1}^{n}(4n - 2)\\ &= (4n + 2)f(n)\\ &= (4n + 2)\frac{(2n)!}{n!}\\ &= 2(2n + 1)\frac{(2n)!}{n!}\\ &= 2(2n + 1)\frac{n+1}{n+1} \frac{(2n)!}{n!}\\ &= \frac{(2n + 2)!}{(n + 1)!}\end{align*}$$

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$$\prod_{i = 1}^n (4i - 2) = \prod_{i = 1}^n 2 \, (2i - 1) = 2^n \, \prod_{i = 1}^n (2i - 1) \frac{\prod\limits_{i = 1}^n 2i}{\prod\limits_{i = 1}^n 2i} = 2^n \frac{\prod\limits_{i = 1}^{2n} i}{2^n \, \prod\limits_{i = 1}^n i} = \frac{(2n)!}{n!}$$

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  • $\begingroup$ Beat me to it. And all the variables are actually in the right places. :) $\endgroup$ – Mike Dec 28 '16 at 11:44
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see this .

Sorry I am not good in latex!!! This way of writing $\frac{2n!}{n!}$ helps in other problems also.

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