Sum of Sequence of Cubes/Proof using Multiplication Table

Proof
We aim to demonstrate that the "Sum of Cubes" is the "Square of the Sum" using simple Multiplication Tables.

On the of the equation, the "Square of the Sum" will be represented by a two dimensional array.

The sum of all numbers contained inside the two dimensional (n x n) array is the "Square of the Sum".

For example:


 * $ \paren {1 + 2}^2 = 1 + 2 + 2 + 4$


 * $\begin{array}{r|cccccccccc}

\paren {1 + 2}^2 & 1 & 2 \\ \hline 1 & 1 & 2 \\ 2  & 2 & 4  \\ \end{array}$


 * $ \paren {1 + 2 + 3}^2 = 1 + 2 + 3 + 2 + 4 + 6 + 3 + 6 + 9$


 * $\begin{array}{r|cccccccccc}

\paren {1 + 2 + 3}^2 & 1 & 2 & 3 \\ \hline 1 & 1 & 2 & 3 \\ 2  & 2 & 4 & 6 \\ 3  & 3 & 6 & 9 \\ \end{array}$


 * $ \cdots $


 * $ \paren {1 + 2 + 3 + \cdots + N}^2 = $


 * $\begin{array}{r|cccccccccc}

\paren {\sum_{i \mathop = 1}^n i}^2 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ 2 & 2 & 4 & 6 & 8 & 10 & 12 & \cdots & 2 N  \\ 3 & 3 & 6 & 9 & 12 & 15 & 18 & \cdots & 3 N \\ 4 & 4 & 8 & 12 & 16 & 20 & 24 & \cdots & 4 N \\ 5 & 5 & 10 & 15 & 20 & 25 & 30 & \cdots & 5 N \\ 6 & 6 & 12 & 18 & 24 & 30 & 36 & \cdots & 6 N \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ N & N & 2 N & 3 N & 4 N & 5 N & 6 N & \cdots & N^2 \\ \end{array}$

From Closed Form for Triangular Numbers:
 * $\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

So the sum of the numbers in the two dimensional n x n array (i.e. "Square of the Sum") is:
 * $\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$

So where is the "Sum of the Cubes"?

Consider the sum of the set of all numbers in each outermost square edge above of length $1$, $2$, $3$, $\ldots$.

We have:

Therefore,

The "Square of the Sum" (i.e. the sum of all of the numbers in an n x n multiplication table) is identical to

The "Sum of the Cubes" (i.e. the sum of the numbers on each incremental outer square-edge is a cube, sum up the outer edges), or:


 * $\forall n \in \Z_{>0}: \ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$