<s>[INST] Summation of product of $m+1$ alternate numbers We know that 
$$\begin{align}
\sum_{r=1}^n \prod_{j=0}^m (r+j)&=\sum_{r=1}^n r(r+1)(r+2)\cdots(r+m)\\
&=(m+1)!\sum_{r=1}^n\binom {r+m}{m+1}\\
&=(m+1)!\binom {n+m+1}{m+2}\\
&=\frac {(n+m+1)^\underline{m+2}}{m+2}\end{align}$$
but is there a similar closed form for 
$$\sum_{r=1}^n \prod_{j=0}^m (r+2j)=\sum_{r=1}^n r(r+2)(r+4)\cdots (r+2m)$$
?

Edit 1
Here's a summary of results from wolframlpha for the first few values of $m$ (please excuse the messy alignment).

From the summary we observe some interesting patterns as follows:


*

*all results have factors $n,(n+1)$

*all results have a factor of $\frac 1{2(m+2)}$ or $\frac 1{m+2}$ where the product of the other factors results in a $2n^{m+2}$ term or $n^{m+2}$ term respectively

*for $m=2$: $(n+4),(n+5)$ are factors

*for $m=4$: $(n+8),(n+9)$ are factors

*for $m=6$: $(n+12),(n+13)$ are factors

*for $m=8$: $(n+16),(n+17)$ are factors

*this implies that for even $m$: $(n+2m), (n+2m+1)$ are factors


Perhaps the observations above might be helpful in working out the solution.
The solution could then possibly be of the form:
$$\begin{cases}
\dfrac {n(n+1)(n+2m)(n+2m+1)\cdot f(n,m)}{2(m+2)}\qquad \text{for even $m$}\\
\dfrac {n(n+1)\cdot g(n,m)}{2(m+2)}\qquad \qquad\qquad\qquad\qquad\text {for odd $m$}
\end{cases}$$ 
 [/INST] Simply for my result, I have:
$$\sum_{r=1}^n\prod_{j=0}^m(r+2j)=S(n,m)=\frac{1}{2m+4}\left(\prod_{k=0}^m(2k+1)+\prod_{k=0}^{m+1}(n+2k)+\prod_{k=0}^{m+1}(n+2k-1)\right)$$
That's a polynomial of $n$ with $m+2$ degrees.
The first few values of $m$
\begin{align*}S(n,1)&=\frac{1}{6}\left(\prod_{k=0}^1(2k+1)+\prod_{k=0}^2(n+2k)+\prod_{k=0}^2(n+2k-1)\right) \\ &=\frac{1}{6}\bigg(1.3+n(n+2)(n+4)+(n-1)(n+1)(n+3)\bigg)\\ &=\frac{1}{6}n(n+1)(2n+7)\end{align*}
\begin{align*}S(n,2)&=\frac{1}{8}\bigg(1.3.5+n(n+2)(n+4)(n+6)+(n-1)(n+1)(n+3)(n+5)\bigg) \\ &=\frac{1}{8}\cdot2n(n+1)(n+4)(n+5)\end{align*}
etc..,
I think so don't closed form for this sum, It is a similarity with the sum $\displaystyle \sum_{k=1}^n k^r$
</s>