Intuitive reason for why $\left(\displaystyle\sum_{i=0}^n i\right)^2 = \displaystyle\sum_{i=0}^n i^3$ [duplicate]

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.

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

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

Is that convincing?

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

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

$$n^3=\frac{n^2(n+1)^2-n^2(n-1)^2}4=S_n-S_{n-1}.$$

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.

• 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?? Aug 30, 2017 at 9:33
• @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.
– user65203
Aug 30, 2017 at 9:37