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[Community Question] Calculus: How is $\left(\frac{n!}{(2n +1)!!}\right)^2 4^n = \left(\frac{(2n)!!}{(2n +1)!!}\right)^2 ?$

One of our user asked:

How is $$\left(\frac{n!}{(2n +1)!!}\right)^2 4^n = \left(\frac{(2n)!!}{(2n +1)!!}\right)^2 ?$$

Could anyone explain this for me please?


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