# The first row of squares for general k

## The first row of squares for general k

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

 (1)

As we mentioned in the main article, there’s a formula for the sum of the first integers:

 (2)

Substituting this into the left part of expression (1) gives

 (3)

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

 (4)

Factoring out the transforms this sum to

 (5)

Now the second bracket here is equal to , the triangular number. Therefore our sum (5) is equal to

as we claimed.