Is there any intuitive reason or deeper meaning to why the following equality holds?

$$\left(\sum_{i=0}^n i\right)^2 = \sum_{i=0}^n i^3$$

I'm not looking for a proof of this, I'm looking for some explanation for this equality if it isn't just a coincidence.

  • 3
    $\begingroup$ For more intuitive answers, look at math.stackexchange.com/questions/61482/… $\endgroup$
    – alans
    Aug 4, 2016 at 12:14
  • 1
    $\begingroup$ There is no intuition for this. Indeed the image posted by @RobertZ is not intuitive at all. $\endgroup$
    – Masacroso
    Aug 4, 2016 at 12:19

2 Answers 2


Take a look at this picture (a nice visual proof):

enter image description here

Is that convincing?

  • 3
    $\begingroup$ This was exactly what i was looking for. Things like this are what I like the most about maths. Thanks $\endgroup$ Aug 4, 2016 at 12:03
  • $\begingroup$ @ZiadFakhoury You Can find more like this in "Proof Without Words" Books... $\endgroup$
    – MR_BD
    Aug 4, 2016 at 12:09
  • $\begingroup$ nice answer @robert $\endgroup$ Dec 12, 2016 at 8:35

From the easy identity $(n+1)^2-(n-1)^2=4n$, we deduce


By telescoping, the sum of consecutive cubes is always a difference of squares.

In particular, the sum of the first cubes is always a perfect square.

Then, it should be intuitive that the sum of the $p^{th}$ powers of the integers are polynomials of degree $p+1$, and comparing to integrals, the leading term must be

$$\sum_{i=1}^n i^p\sim\frac{n^{p+1}}{p+1}.$$

So, looking for a case such that $$\left(\frac{n^{p+1}}{p+1}\right)^2=\frac{n^{q+1}}{q+1}$$

we find the only solution

$$\left(\frac{n^2}2\right)^2=\frac{n^4}4$$ giving identical asymptotic behaviours.

Combining the previous two properties, the two sequences must coincide exactly.

  • $\begingroup$ dear prof. Yves Daoust can you help me for ask Math question in stack.exchange??..Because this is so hard question for ask..And I can not turn the question into mathematics language..(+latex and englosh language)..Please..can you help me as your Little student.. :( I prepared direk link.. Can you look?? $\endgroup$ Aug 30, 2017 at 9:33
  • 2
    $\begingroup$ @MathLife: enter your question as usual and I'll take a look. But I am not doing the LaTeX encoding for you. You need to learn. $\endgroup$
    – user65203
    Aug 30, 2017 at 9:37

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