via Sum of Sequence of Squares – ProofWiki.
But, why would you want to know what the sum of the square numbers is in the first instance.
Mathematics often driven by the need to calculate lengths, areas and volumes. Calculating lengths gets more complicated when they slope, the areas of shapes gets more complicated when they are triangular and the equivalent in three dimensions would have to be a pyramid.
Imagine a triangle created by a series of identical squares of side s, one at the top, sitting on two, sitting on three and so on. We could approximate the area of the triangle by adding up the area o all of the squares used to construct it. This will lead us to needing to know the sum of all the integers.
area of the first n rows is equal to
A=s^2 + 2 s^2 + 3 S^2 + 4 s^2 + … + (n-2) s^2 + (n-1) s^2 + n s^2
A=s^2 ( 1 + 2 + 3 + 4 + … + (n-2) + (n-1) + n )
The sum of the first n integers is n (n+1) / 2
So the area of the triangle is equal to
A= s^2 n (n+1) / 2
We can modify our original square of side s to a rectangle to cope with triangles that are longer than they are high or vice versa but the result will still be the same.
So it may have been that in the first instance someone just decided to find the sum of the first n integers for no other reason than because its the sort of thing a mathematician will just like to investigate, or it may have been a need to find the area of a slightly more complex shape. It could be a mixture of both.
Transferring this problem to one in three dimensions using the same system will lead to the need of calculating the sum of the square numbers. The sum of the first n square numbers is a lot more complicated to discover (hence the link above) but the end result is
Sum of first n squares = n (2 n + 1) (n + 1) / 6
Engineering 