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I don't know why this is true:

$$(2k+1)!!=\frac{(2k+1)!}{2^kk!}$$

Can anyone explain it for me? I know what is double factorial, but would like to know how this formula was derived. Thanks.

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    $\begingroup$ If $x!! = (x!)!$, then this is certainly not true, since $(2k+1)!! > (2k+1)!>\frac{(2k+1)!}{2^{k}k!}$. $\endgroup$ – Tlön Uqbar Orbis Tertius Sep 13 '14 at 9:51
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    $\begingroup$ @HerbertQuain. Hummmm $(2k+1)!! > (2k+1)!$ does not seem correct to me $\endgroup$ – Claude Leibovici Sep 13 '14 at 9:54
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    $\begingroup$ Okay, my bad. The author might define this symbol in his post, then. $\endgroup$ – Tlön Uqbar Orbis Tertius Sep 13 '14 at 9:54
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    $\begingroup$ @JackD'Aurizio I learned something again. Btw, couldn't they think of something better? Very confusing. $\endgroup$ – drhab Sep 13 '14 at 9:55
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    $\begingroup$ @HerbertQuain apologize? I think that the inventors of this notation owe you an apology. $\endgroup$ – drhab Sep 13 '14 at 9:59
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HINT : $$(2k+1)!!=1\cdot 3\cdot 5\cdot \cdots (2k-1)\cdot (2k+1)$$ $$(2k+1)!=1\cdot 2\cdot 3\cdot \cdots (2k)\cdot (2k+1) $$ $$2^k\cdot k!=2^k\times \{1\cdot 2\cdot 3\cdot \cdots (k-1)\cdot k\}=2\cdot 4\cdot 6\cdot\cdots (2k-2)\cdot (2k).$$

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By multiplying both sides by $(2k)!! = (2k)(2k-2)\cdot\ldots\cdot 2 = 2^k\cdot k!$, you get: $$(2k+1)! = (2k+1)!.$$

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