geometric proof that the sum of the cubes is equal to the square of the sum

It's still too hard for me to appreciate the real value of your achievement, though, is this related to some kind of the sense of humour?
no. it's a real proof. but i didn't explain properly what of, hence it's really just a pretty picture.

1^3 + 2^3 + 3^3 +...+ n^3 = (1 + 2 + 3 +...+ n)^2

Here n^k means n to the power of k. It's a formula for the sum of the first n cube numbers: 1*1*1 + 2*2*2 +...+ n*n*n
Each cube number is represented by n squares, each of size n*n. All the squares the same size are in the same colour, alternating dark and light, and one of the squares is cut in half for each even number. So the little square in the top left has area 1 = 1*1 = 1*1*1. The strip of light and dark purple squares along the bottom and right consists of 7 squares, each of dimensions 7*7, giving total area 7 cubed. There are 6 squares of side length 6 in the adjacent blue strip, but one of them has been cut in half, with a piece at either end of the strip.
So the total area is 1*1*1 + 2*2*2 +...+ 7*7*7. It's easy to see that the area is also (1 + 2 +...+7)^2, since the length of one side of the whole square is 1 + 2 +...+ 7.
Hey, that's nifty.
i aim to please :)
And the math proof?
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