The double factorial, \(n!!\), is defined as the product of all positive integers less than or equal to \(n\) that have the same parity as \(n\).
Some examples:
When reading about it in a book authored by a friend of mine, the identity \((2n)!! = 2^n n!\) was also presented, and I'll prove it now by induction. For \(n = 1\), we have \(2!! = 2 = 2^1 \times 1!\), which is true. Now, assuming the identity holds up to \(n\), we show it holds for \(n + 1\):
\[ \begin{align} (2(n + 1))!! &= (2(n + 1)) \times (2n)!! \\ &= (2(n + 1)) \times 2^n n! \\ &= (n + 1) \times 2^{n + 1} n! \\ &= 2^{n + 1} (n + 1)! \end{align}\]
Done!
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