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Could someone please help me as to how I'd go about evaluating:

$$\prod_{r=1}^{n} (2r+1)$$

I have that written out, it is:

$$1 \cdot 3 \cdot 5 \cdots (2n-1) \cdot (2n+1)$$

furthermore:

$$\prod_{r=1}^{2n+1} r = 1 \cdot 2 \cdot 3 \cdots (2n-1) \cdot 2n \cdot (2n+1) = (2n+1)!$$

which looks similar but from there I'm stuck :(

Any help is much appreciated!

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  • $\begingroup$ learn about $n!!$ please $\endgroup$ – eccstartup Jun 20 '13 at 15:30
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This is usually denoted as a double factorial: $$ (2n+1)!!:=\prod_{k=0}^n(2k+1) $$ and $$ (2n)!!:=\prod_{k=1}^n2k = 2^nn! $$ so that $$ (2n+1)!! = \frac{(2n+1)!}{(2n)!!} = \frac{(2n+1)!}{2^nn!}. $$

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$2 \cdot 4 \cdot 6 \ldots 2n$ can be written as $2^n (1 \cdot 2 \cdot 3 \ldots n)=2^n n!$

Thus, you can write your product as:

$$\frac{(2n+1)!}{2^n n!}$$

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