Here is an interesting relationship:
$$\sum_{r=1}^n r(r+m)(r+2m)=\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m} r\right)\tag{1}$$ $$\text{i.e.} \;\; 1\cdot(1+m)(1+2m)+2\cdot (2+m)(2+2m)+3\cdot (3+m)(3+2m)+\cdots +n(n+m)(n+2m)=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+2m))$$ which gives the following for the first few values of $m$: $$\scriptsize\begin{align} &1\cdot 1\cdot 1+2\cdot 2\cdot2+3\cdot3\cdot3+\cdots+n\cdot n\cdot n&&=(1+2+3+\cdots+n)(1+2+3+\cdots+n)\\ &1\cdot2\cdot3+2\cdot3\cdot4+3\cdot5\cdot7+\cdots+n(n+1)(n+2)&&=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+2))\\ &1\cdot3\cdot5+2\cdot4\cdot6+3\cdot5\cdot7+\cdots+n(n+2)(n+4)&&=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+4))\\ \end{align}$$
This can be proven by expanding both sides and finding that the result is $$\frac {n(n+1)}2\cdot \frac {(n+2m)(n+2m+1)}2\tag{2}$$
Question
Can the result can be proven without expansion to the closed form but only by manipulating summands and limits, i.e. prove $(1)$ without first expanding to $(2)$?
Some interesting but commonly-known results follow from this.
Setting $m=0$ gives the "sum of cubes as square of sum of integers" result: $$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2$$
Setting $m=1$ gives the "discrete integral"
$$\sum_{r=1}^n r^\overline{3}=\frac{\;n^\overline{4}}4$$ where $r^\overline{a}$ is the symbol for the rising factorial.
Sum[ Product[r + k*m, {k, 0, 3}], {r, 1, n}]
change the 3 for the closed form for various other ones. I suppose i'm not doing anything helpful, since i'm not actually answering the question. I feel the answer is just fancy combinatorics. $\endgroup$