Let $k$ 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 $k$ blue. First, we are going to prove that the numbers within the $nth$ larger square from the left add to $$(2n-1)T_{k-1}^2,$$ where $T_{k-1}$ is the $(k-1)st$ triangular number. Let's call the square in question $S.$ First note that the numbers within the top row of $S$ are $$1+k(n-1),2+k(n-1),3+k(n-1),...,((k-1)+k(n-1)).$$ Adding those up we get the sum $$\text{Missing left or extra right}$$ which we can rearrange to give $$[1+2+...+(k-1)]+[(k-1)k(n-1)].$$ As we mentioned in the main article, there's a formula for the sum of the first $k-1$ integers: $$1+2+...+(k-1)=\frac{(k-1)k}{2}.$$ Substituting this into the left part of expression (1) gives $$\text{Missing begin{equation} or extra end{equation}}$$ Now the $(k-1)st$ triangular number is equal to the sum of the first $k-1$ integers, so it's also equal to the right hand side of equation (2). This means that expression (3) is equal to $$(2n-1)T_{k-1}.$$ Now notice that the numbers in the second row of $S$ are 2 times the numbers in the first row, so their sum is equal to $$2(2n-1)T_{k-1}.$$ The numbers in the second row of $S$ are 3 times the numbers in the first row, so their sum is equal to $$3(2n-1)T_{k-1}.$$ We can continue in this vein until we come to the sum of the last row of $S$, which is equal to $$(k-1)(2n-1)T_{k-1}.$$ Adding up the sums of all the rows in $S$ therefore gives $$(2n-1)T_{k-1}+2(2n-1)T_{k-1}+...+(k-1)(2n-1)T_{k-1}.$$ Factoring out the $(2n-1)T_{k-1}$ transforms this sum to $$(2n-1)T_{k-1}(1+2+3+...+(k-1)).$$ Now the second bracket here is equal to $T_{k-1}$, the $(k-1)st$ triangular number. Therefore our sum (5) is equal to $$(2n-1)T_{k-1}^2,$$ as we claimed.

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