How to write $1 \times 3 \times 5 \times \cdots \times (2n-1)$ as a sequence? I am a second year IB Mathematics HL student and I am trying to figure out how to write the equation for the following sequence:
1×3×5×...×(2n-1)
I’m pretty sure it involves factorials, but (2n-1)! doesn’t work, and neither does 2(n!)-1, obviously.
If you know how to do this or can suggest another approach, I would really appreciate your help!  Thank you!
 A: You have a notation for the multifactorials, which takes into account only one factor on $k$. 
For instance, the double factorial is defined recursively by the relation
$$\begin{cases}n!!= 1&\text{if }-2<n\le 0,\\ n\cdot (n-2)!!&\text{if } n>0.\end{cases}
$$
Thus,
$$(2n-1)!!=1\cdot 3 \cdot 5\cdots (2n-1),\qquad (2n)!!=2\cdot 4\cdot6\cdots 2n.$$
Also, note that $ \; n!=n!! \, (n-1)!!$.
Similarly, you have a triple factorial, defined by
$$n!!!=\begin{cases} 1&\text{if }-3<n\le 0,\\ n\cdot (n-3)!!!&\text{if } n>0.\end{cases}
$$
A: This is a good exercise
in using product notation.
$\begin{array}\\
f(n)
&=1\ 3\ ...\ (2n-1)\\
&=\prod_{k=1}^n (2k-1)
\qquad\text{This is the product of the first $n$ odd numbers}\\
&=\dfrac{\prod_{k=1}^n (2k-1)\prod_{k=1}^n (2k)}{\prod_{k=1}^n (2k)}
\qquad\text{join the even numbers to the odd ones}\\
&=\dfrac{\prod_{k=1}^{2n} k}{2^n\prod_{k=1}^n k}
\qquad\text{numerator is the product of the first $2n$ numbers}\\
&=\dfrac{(2n)!}{2^nn!}
\qquad\text{renaming product of first consecutive numbers with factorials}\\
\end{array}
$
A: You may use Π(2k-1) for k=1 to k=n. The upper-case letter Π is used as a symbol for: The product operator in mathematics, indicated with capital pi notation ∏ (in analogy to the use of the capital Sigma Σ as summation symbol). 
