# Intuitive ways to get formula of cubic sum

Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$ I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$$ But how to prove it? Please help me. Grazie!

• You have a lot of interesting questions for an eighth grader! Jul 24, 2014 at 13:55
• @JasonKnapp I'm just wondering because I tried trial & error during doing my homework and it works Jul 24, 2014 at 13:58
• In general given any arbitrary sequence there is a very efficient general method to find a polynomial closed form. Feb 11, 2019 at 7:44

We have $$\sum_{k=1}^{n}k^3 = 1 + 8 + 27 + \ldots + n^3 = \\ \underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \ldots = \\ \underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \ldots$$ which is $$\left( \sum_{k=1}^{n}k \right)^2$$

• This is also very nice! Thanks! Jul 24, 2014 at 14:05
• Yes, I find it quite amusing, just messing around with numbers and see what comes out =) Jul 25, 2014 at 8:34

Source.

or this one which is clearer:

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• Very nice! Thanks for your answer :) Jul 24, 2014 at 14:01
• @LLawliet You're welcome, glad I could help! ;) Jul 24, 2014 at 14:01

Here are some 'Proof without Words' for this identity;

As Hakim showed (but this one might be slightly clearer);

Here's another very clean illustration by Anders Kaseorg;

Here, the total top volume is undoubtedly $$1^2+2^2+3^3+\cdots +n^2,$$ while on the volume of the bottom part is; $$1\left(1+2+\cdots+n\right)+2\left(1+2+\cdots+n\right)+\cdots +n\left(1+2+\cdots+n\right)=\left(1+2+\cdots+n\right)^2.$$

Here's a proof by induction I came up with about two years ago.

Let $$s_k(n) =\sum_{i=1}^n i^k$$. Want to show that $$s_3(n) = (s_1(n))^2$$.

If you want to prove by induction that $$a(n) = b(n)$$, there are two possibilities: (1) Show $$a(0)-b(0) = 0$$ and $$a(n)-b(n) = 0 \implies a(n+1)-b(n+1) = 0$$; (2) show $$a(0) = b(0)$$ and $$a(n) = b(n) \implies a(n+1)-a(n) =b(n+1)-b(n)$$.

These are, of course, equivalent, but I find that (2) is often easier since the math is simpler in the initial parts.

In this case, $$s_3(n+1)-s_3(n) =(n+1)^3 =n^3+3n^2+3n+1$$ and

$$(s_1(n+1))^2-(s_1(n))^2 =(s_1(n)+n+1)^2-(s_1(n))^2\\ =s_1^2(n)+2(n+1)s_1(n)+(n+1))^2-(s_1(n))^2\\ =2(n+1)s_1(n)+n^2+2n+1\\$$

so we want to show that $$n^3+3n^2+3n+1 =2(n+1)s_1(n)+n^2+2n+1$$ or $$n^3+2n^2+n =2(n+1)s_1(n)$$ or $$n(n^2+2n+1) =2(n+1)s_1(n)$$ or $$n(n+1)^2 =2(n+1)s_1(n)$$ or $$s_1(n) =\dfrac{n(n+1)}{2}$$.

This can either be assumed known or, in turn, proved by induction using $$\dfrac{(n+1)(n+2)}{2} -\dfrac{n(n+1)}{2} =\dfrac{n^2+3n+2-(n^2+n)}{2} =\dfrac{2n+2}{2} =n+1$$.