Problem: Find the sum of the first 1000 natural numbers.
Solution: Write the numbers twice as follows:
Summing the numbers in each column, we have:
And dividing by two we have
This method clearly works not just for 1000, but any natural number .
Problem: Find the sum of the squares of the first natural numbers
Solution: Arrange the numbers three times, in triangles as follows:
Here we prove what is obvious from the picture: that the numbers in the “sum” triangle work out as expected.
In the leftmost triangle, there are copies of in the th row (indexing from 1), so the sum of the th row is , and the sum of all the numbers in the triangle is . Since the other two “summand” triangles are just rotations of the first triangle, their sums are also the sum of the first squares.
In the “sum” triangle at the th row spaces from the left (indexing both from zero), we have the value . In each triangle there are different numbers (as per the first proof), so there are that many copies of in the fourth triangle. Summing up the fourth triangle gives three times our desired result, and hence
This second proof was shown to me by Sándor Dobos, a professor of mine in Budapest whom I remember fondly.