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

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

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!

• learn about $n!!$ please – eccstartup Jun 20 '13 at 15:30

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 $$(2n+1)!! = \frac{(2n+1)!}{(2n)!!} = \frac{(2n+1)!}{2^nn!}.$$
$2 \cdot 4 \cdot 6 \ldots 2n$ can be written as $2^n (1 \cdot 2 \cdot 3 \ldots n)=2^n n!$
$$\frac{(2n+1)!}{2^n n!}$$