Possible equation for cubes? $N + 6\sum_{n = 1}^N \frac{n(n-1)}{2} = N^3$

Question

I recently discovered an equation that can represent cubes:

$$N + 6\sum_{n = 1}^N \frac{n(n-1)}{2} = N^3$$

I don't know how to prove it but it is accurate for all values of $N$ that I try. So my questions are:

Is it accurate?

If yes, then

Has it been identified before?

How could it be used in mathematics?

Answer 1

$N + 6\sum_{n = 1}^N \frac{n(n-1)}{2}$

$N + 3\sum_{n = 1}^N {n(n-1)} $

$N+3\sum_{n = 1}^N n^2-n=N+3\bigg(\frac{(N)(N+1)(2N+1)}{6}-\frac{(N)(N+1)}{2}\bigg)$

$N+3\sum_{n = 1}^N n^2-n=N+\bigg(\frac{(N)(N+1)(2N+1)}{2}-\frac{3(N)(N+1)}{2}\bigg)$

$N+3\sum_{n = 1}^N n^2-n=N+\bigg(\frac{(N)(N+1)(2N-2)}{2}\bigg)$

$N+3\sum_{n = 1}^N n^2-n=N+\bigg(N(N+1)(N-1)\bigg)$

$N+3\sum_{n = 1}^N n^2-n=N(1+N^2-1)=N^3$

$\text{Hence Proved}$

Answer 2

Here's an inductive proof:

The equation is "trivial" for $ \ N = 1 \ \ , $ since $ \ \sum_{n = 1}^1 \frac{n(n-1)}{2} \ = \ \frac{1·0}{2} \ \ . $ We'll start from $ \ N = 2 \ \ $ for which $ \ 2 \ + \ 6 · \sum_{n = 1}^2 \frac{n(n-1)}{2} \ = \ 2 \ + \ 6 · \left[ \frac{1·0}{2} + \frac{2·1}{2} \right] \ = \ 2 \ + \ 6·1 \ = \ 8 \ = \ 2^3 \ \ . $

We suppose it to be true that $ \ N \ + \ 6·\sum_{n = 1}^N \frac{n(n-1)}{2} \ = \ N^3 \ \ . $ For the next integer $ \ (N+1) \ \ , $ $$ \ (N + 1) \ + \ 6·\sum_{n = 1}^{N + 1} \frac{n(n-1)}{2} $$ $$ = \ \ \left[ \ \underbrace{N \ + \ 6·\sum_{n = 1}^{N} \frac{n(n-1)}{2}}_{N^3} \ \right] \ + \ 1 \ + \ 6·\left[ \ \frac{(N+1)·([N+1]-1)}{2} \ \right] $$ $$ = \ \ N^3 \ + \ 1 \ + \ 3·(N+1)·N \ \ = \ \ N^3 \ + \ 1 \ + \ 3·N^2 \ + \ 3·N \ \ = \ \ (N+1)^3 \ \ . \ \ \blacksquare $$

The fact that $ \ \frac{N·(N-1)}{2} \ $ is a "triangular" number and so $ \ \sum_{n = 1}^{N} \frac{n·(n-1)}{2} $ produces the "tetrahedral" numbers leads me to believe that there may be a "visual" proof ("proof without words"), but I haven't found a reliable one so far.