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 $$\frac {n(n+1)}2\cdot \frac {(n+2m)(n+2m+1)}2\tag{2}$$

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.

  • $\begingroup$ Not what you are looking for, but FWIW the two sides are quadratics in $m$ which you showed match at $2$ points. If you verify the equality at a $3^{rd}$ point e.g. $m=-1$ then that would prove the identity in general. $\endgroup$
    – dxiv
    Jun 14, 2017 at 1:54
  • $\begingroup$ Some quick things I can see, $\sum r \mid \sum r (r+m) (r + 2m) \dots (r+km)$ for $k = 0, 1, 2, 3, 4, 5$, my guess is that it continues in this fashion forever. This might be worth proving? $\endgroup$
    – mdave16
    Jun 14, 2017 at 2:09
  • $\begingroup$ Other wild guesses, if $k$ above is even, like in your case, then the factor follows a similar form to yours above. For $k=4$, it is $\sum_1^n r \sum_1^{4m + n} r \frac{9m^2 + 8mn + 4m + 2n^2 + 2n - 1}{3}$. Similar but more ugly fraction for $k=6$. The fraction is $240m^4 + 312m^3n + 156m^3 + 160m^2n^2 + 160m^2n - 26m^2 + 36mn^3 + 54mn^2 - 6mn + 3n^4 + 6n^3 - n^2 - 4n + 2$ all over $6$. $\endgroup$
    – mdave16
    Jun 14, 2017 at 2:17
  • $\begingroup$ btw, i'm using mathematica to play about with this, 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. $\endgroup$
    – mdave16
    Jun 14, 2017 at 2:41

2 Answers 2


I will use the identity

$$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2$$

It can be proved without direct expansion. (https://math.stackexchange.com/a/1215805/268334)

\begin{align} &\;\sum_{r=1}^n r(r+m)(r+2m)\\ =&\;\sum_{r=1}^n [(r+m)^3-m^2(r+m)]\\ =&\;\sum_{r=1}^{n+m}r^3-\sum_{r=1}^mr^3-m^2\sum_{r=1}^n (r+m)\\ =&\;\left(\sum_{r=1}^{n+m}r\right)^2-\left(\sum_{r=1}^mr\right)^2-m^2\sum_{r=1}^n (r+m)\\ =&\;\left(\sum_{r=1}^{n+m}r-\sum_{r=1}^mr\right)\left(\sum_{r=1}^{n+m}r+\sum_{r=1}^mr\right)-m^2\sum_{r=1}^n (r+m)\\ =&\;\left(\sum_{r=1}^n (r+m)\right)\left(\sum_{r=1}^{n+m}r+\sum_{r=1}^mr\right)-m^2\sum_{r=1}^n (r+m)\\ =&\;\left(\sum_{r=1}^n (r+m)\right)\left(\sum_{r=1}^{n+m}r+\sum_{r=1}^mr-m^2\right)\\ =&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-\sum_{r=1}^m(r+m+n)+\sum_{r=1}^mr-m^2\right)\\ =&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-m(m+n)-m^2\right)\\ =&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-\sum_{r=1}^m(r+m+n)+\sum_{r=1}^mr-m^2\right)\\ =&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-m(2m+n)\right)\\ =&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)+mn\sum_{r=1}^{n+2m}r-m(2m+n)\sum_{r=1}^nr-m^2n(2m+n)\\ =&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)+\frac{1}{2}mn(n+2m)(n+2m+1)\\ &\qquad-\frac{1}{2}m(2m+n)n(n+1)-m^2n(2m+n)\\ =&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)+\frac{1}{2}mn(n+2m)(n+2m+1-n-1-2m)\\ =&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right) \end{align}

  • $\begingroup$ Thanks for your solution (+1). $\endgroup$ Jun 14, 2017 at 17:27
  • $\begingroup$ @hypergeometric But I believe there must be a more beautiful way to prove it. $\endgroup$
    – CY Aries
    Jun 14, 2017 at 17:29

Consider the summation $$\sum_{i=1}^si \sum_{j=1}^{s+2m}j$$ When $s=r-1$, the summation is equal to $$\big(1+2+\cdots+(r-1)\big)\big(1+2+3+\cdots+(r-1+2m)\big)$$ When $s=r$, the summation is equal to $$\big(1+2+\cdots+(r-1)+r\big)\big(1+2+3+\cdots+(r-1+2m)+(r+2m)\big)$$ New terms arising from putting $s=r$ are:

$$\require{cancel}\begin{align} &\quad r\sum_{i=1}^{r+2m}i&&+(r+2m)\sum_{j=1}^rj&&-r(r+2m)\\ &=r\sum_{i=0}^{r+2m}i&&+(r+2m)\sum_{j=0}^rj&&-r(r+2m)\tag{1}\\ &=r\sum_{i=0}^{r+2m}\big(r+2m-i\big)&&+(r+2m)\sum_{j=0}^r \big(r-j\big)&&-r(r+2m)\tag{2}\\ &=\frac 12 r\sum_{i=0}^{r+2m}(r+2m)&&+\frac 12 (r+2m)\sum_{j=0}^r r&&-r(r+2m)\tag{(1)+(2))/2}\\ &=\frac 12 r(r+2m)\sum_{i=0}^{r+2m}1&&+\frac 12 r(r+2m)\sum_{j=0}^r 1&&-r(r+2m)\\ &=\frac 12 r(r+2m)\sum_{i=1}^{r+2m}1&&+\frac 12 r(r+2m)\sum_{j=1}^r 1&&\cancel{-r(r+2m)}\\ &\; + \cancel{\frac 12 r(r+2m)}&&+\cancel{\frac 12 r(r+2m)}\\ &=r(r+2m)\frac {r+2m}2&&+r(r+2m)\frac r2\\ &=r(r+m)(r+2m) \end{align}$$ Summing from $r=1$ to $n$ gives

$$\color{red}{\sum_{r=1}^n r\sum_{r=1}^{n+2m}r=\sum_{r=1}^n r(r+m)(r+2m)}$$


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