TITLE: Evaluating $\prod_{r=1}^{n} (2r+1)$
QUESTION [5 upvotes]: 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!

REPLY [5 votes]: 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 
$$
  (2n+1)!! = \frac{(2n+1)!}{(2n)!!} = \frac{(2n+1)!}{2^nn!}.
$$

REPLY [4 votes]: $2 \cdot 4 \cdot 6 \ldots 2n$ can be written as $2^n (1 \cdot 2 \cdot 3 \ldots n)=2^n n!$
Thus, you can write your product as:
$$\frac{(2n+1)!}{2^n n!}$$