# Sum of Sequence of Cubes/Visual Demonstration

## 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 $n = 5$:

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.