Sum of Sequence of Cubes/Visual Demonstration

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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

A visual illustration of the proof for $n = 5$:


SumOfCubes-Floyd.png


The number of squares of side $n$ is seen to be $4 n$.


To go from $n$ to $n + 1$, a ring of $4 \paren {n + 1}$ squares of side $n + 1$ is to be added:

$4$ for the one at each corner
$4 n$ for the ones that abut the sides of the $n + 1$ squares of side $n$.


The length of one side is given by:

$S = 2 \paren {1 + 2 + \cdots + n}$


The length of one side is also given by:

$S = n \paren {n + 1}$


The area is therefore given in two ways as:

$A = 4 \paren {1 + 2 + \cdots + n}^2 = \paren {n \paren {n + 1} }^2$


and also as:

\(\ds A\) \(=\) \(\ds 4 \times 1^2 + 4 \times 2 \times 2^2 + 4 \times 3 \times 3^2 + \cdots + 4 \times n \times n^2\)
\(\ds \) \(=\) \(\ds 4 \paren {1^3 + 2^2 + 3^3 + \cdots + n^3}\)

The result follows by equating the expressions for area.

$\blacksquare$


Historical Note

This visual demonstration of the Sum of Sequence of Cubes was reported by Solomon W. Golomb as having been devised by Warren Lushbaugh.


Sources