Let be a positive integer. Look at the first row of larger squares in the complement of the blue grid you get from colouring squares corresponding to multiples of blue. First, we are going to prove that the numbers within the larger square from the left add to
where is the triangular number.
Let’s call the square in question First note that the numbers within the top row of are
Adding those up we get the sum
which we can rearrange to give
As we mentioned in the main article, there’s a formula for the sum of the first integers:
Substituting this into the left part of expression (1) gives
Now the triangular number is equal to the sum of the first integers, so it’s also equal to the right hand side of equation (2). This means that expression (3) is equal to
Now notice that the numbers in the second row of are 2 times the numbers in the first row, so their sum is equal to
The numbers in the second row of are 3 times the numbers in the first row, so their sum is equal to
We can continue in this vein until we come to the sum of the last row of , which is equal to
Adding up the sums of all the rows in therefore gives
Factoring out the transforms this sum to
Now the second bracket here is equal to , the triangular number. Therefore our sum (5) is equal to
as we claimed.