# Sum of Cubes

December 23, 2009 2 Comments

It is well known that:

(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:

(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.

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