# Sum of Cubes

It is well known that:

$1+2+3+\cdots+n = \frac{1}{2}n(n+1)$

(The usual way to prove it is to consider twice the quality on the left hand side and bracket terms so that each one equals n+1. Alternatively one can sum k²-(k-1)² in two ways – telescoping and expanding.)

It is also the case that:

$1^3+2^3+3^3+\cdots+n^3 = \left(\frac{1}{2}n(n+1)\right)^2$

(It is simple to prove this by summing a similar telescoping sum as above but with forth powers.)

However, this does not explain why one is the square of the other! The following picture geometrically shows it, which is perhaps more intuitive:

To see the identity, one must understand this picture in two ways. For the first way, note that the black square has size 1, the red squares have size 2×2 and their are 2 of them, the yellow squares have size 3×3 and there are 3 of them, the green squares have size 4×4 and there are 4 of them, and so on.

Thus the area of the square must be the sum of the first 7 cubes. If the picture were bigger it would be the sum of the first however-many cubes.

However, the area is also the length squared, which can be seen to be the sum of the first 7 numbers.

I cannot remember where this image comes from, however I did not draw it myself. Apparently the argument was printed in The Art Of Programming though.