Turning product sequences into factorials

Question

I am trying to figure out the steps between these equal expressions in order to get a more general understanding of product sequences: $$\prod_{k=0}^{n}\left(3n-k\right) + \prod_{k=n}^{2n-3}\left(2n-k\right) = \prod_{j=2n}^{3n}j + \prod_{j=3}^{n}j =\frac{(3n)!}{(2n-1)!}+\frac{n!}{2}$$

I know that $ n! :=\prod_{k=1}^{n}k$ but I can't figure out how that helps me understand the above equation.

edit: Thank you for the great help! Another thing I don't understand, is how I get from $\prod_{k=0}^{n}\left(3n-k\right) + \prod_{k=n}^{2n-3}\left(2n-k\right)$ to $\prod_{j=2n}^{3n}j + \prod_{j=3}^{n}j$. Any help with understanding this is much appreciated, I will try to figure it out myself while I wait for answers.

Answer 1

Let's evaluate a few terms for $$\prod_{k=0}^n (3n-k)$$

when $k=0$, we are multipliying $3n$.

when $k=1$, we are multipliying $3n-1$.

when $k=2$, we are multipliying $3n-2$.

and so on. Each time, the individual term reduces by $1$.

when $k=n$, we are multiplying $3n-n=2n$

Hence we are just multiplying every term from $2n$ to $3n$.

$$\prod_{k=0}^n (3n-k)=\prod_{j=2n}^{3n}j=\frac{(3n)!}{(2n-1)!}$$

Answer 2

We obtain for integers $n\geq 2$:

\begin{align*} \color{blue}{\prod_{k=0}^{n}}&\color{blue}{\left(3n-k\right) + \prod_{k=n}^{2n-3}\left(2n-k\right)}\\ &=\prod_{k=0}^n(2n+k)+\prod_{k=n}^{2n-3}(k+3-n)\tag{1}\\ &\,\,\color{blue}{=\prod_{k=2n}^{3n}k+\prod_{k=3}^{n}k}\tag{2}\\ &=\left(\prod_{k=1}^{3n}k\right)/\left(\prod_{k=1}^{2n-1}k\right)+\left(\prod_{k=1}^n k\right)/\left(\prod_{k=1}^2k\right)\tag{3}\\ &\,\,\color{blue}{=\frac{(3n)!}{(2n-1)!}+\frac{n!}{2}} \end{align*}

Comment:

• In (1) we change in both products the order of multiplication: $k\to (n-k)$ and $k\to ((2n-3)-k+n)$.

• In (2) we shift the left product by $2n$ via $k\to2n+k$ and we shift the right product by $-n+3$ via $k\to k+n-3$.

• In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.

We start with $n\geq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.