How to write $1 \times 3 \times 5 \times \cdots \times (2n-1)$ as a sequence?

Question

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!

Answer 1

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 $$n!!!=\begin{cases} 1&\text{if }-3<n\le 0,\\ n\cdot (n-3)!!!&\text{if } n>0.\end{cases} $$

Answer 2

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).

Answer 3

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} $