The first row of squares for general k

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

$1+k(n-1), 2+k(n-1), 3+k(n-1), ... , \left((k-1)+k(n-1)\right).$

Adding those up we get the sum

$1+k(n-1)+2+k(n-1)+3+k(n-1)+...+(k-1)+k(n-1)\right),$

which we can rearrange to give

$\begin{equation} \left[1+2+...+(k-1)\right] + \left[(k-1)k(n-1)\right].\end{equation}$

(1)

As we mentioned in the main article, there’s a formula for the sum of the first $k-1$ integers:

$\begin{equation} 1+2+...+(k-1) = \frac{(k-1)k}{2}. \end{equation}$

(2)

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

$\begin{equation} \frac{(k-1)k}{2} + (k-1)k(n-1) = \frac{(k-1)k+2(k-1)(n-1)k}{2}=\frac{(k-1)k\left(1+2(n-1)\right)}{2}=\frac{(k-1)k}{2}\left(2n-1).\end{equation}$

(3)

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

$\begin{equation} (2n-1)T_{k-1}+2(2n-1)T_{k-1}+ ... +(k-1)(2n-1)T_{k-1}.\end{equation}$

(4)

Factoring out the $(2n-1)T_{k-1}$ transforms this sum to

$\begin{equation} (2n-1)T_{k-1}\left(1+2+3+...+(k-1)\right).\end{equation}$

(5)

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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