Sum of Sequence of Cubes/Proof using Multiplication Table

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Theorem

$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$


Proof

$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \paren {1 + 2 + 3 + \cdots + N}^2$
\(\ds \paren {1 + 2 + 3 + \cdots + N}^2\) \(=\) \(\ds 1 \times \paren {1 + 2 + 3 + \cdots + N}\)
\(\ds \) \(+\) \(\ds 2 \times \paren {1 + 2 + 3 + \cdots + N}\)
\(\ds \) \(+\) \(\ds \cdots\)
\(\ds \) \(+\) \(\ds N \times \paren {1 + 2 + 3 + \cdots + N}\)

Organizing the terms above in a square matrix:

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


From Closed Form for Triangular Numbers:

$\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Therefore:

$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$

Next we notice that the sum of the terms in row N with the terms in column N (less $N^2$ so as not to double count) is precisely $N^3$.

\(\ds 1^3\) \(=\) \(\ds 1\)
\(\ds 2^3\) \(=\) \(\ds 2 + 4 + 2\)
\(\ds 3^3\) \(=\) \(\ds 3 + 6 + 9 + 6 + 3\)
\(\ds \) \(\cdots\) \(\ds \)
\(\ds n^3\) \(=\) \(\ds n + 2 n + 3 n + \cdots + \paren {n - 1} \paren n + n^2 + \paren {n - 1} \paren n + \cdots + 2 n + n\) 1+2+...+n+(n-1)+...+1 = n^2

Therefore:

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

$\blacksquare$