Any large square for general k

To warm up, let’s start with $k=3$ . We want to prove that the numbers within the large square that’s $nth$ from the left ad $mth$ from the top sum to

$(2m-1)(2n-1)T_3^2= (2m-1)(2n-1)9.$

First look at the large square that’s $nth$ from the left and lies in the first row of large squares. As we already noted above, the two numbers in the top row of this white square are $3(n-1)+1=3n-2$ and $3(n-1)+2=3n-1.$

By some very similar reasoning we can work out that the top two numbers in the large square that’s $nth$ from the left and $mth$ from the top are

$\left(3(m-1)+1\right)\left(3n-2\right)=(3m-2)(3n-2)$

and

$\left(3(m-1)+1\right)\left(3n-1\right)=(3m-2)(3n-1)$

Similarly, the bottom two numbers in the $nth$ white square from the left in the $mth$ row of white squares are

$\left(3(m-1)+2\right)\left(3n-2\right)=(3m-1)(3n-2)$

and

$\left(3(m-1)+2\right)\left(3n-1\right)=(3m-1)(3n-1).$

Adding these four numbers gives

$\begin{array}{cc}& (3m-2)(3n-2) \\ + & (3m-2)(3n-1) \\ + & (3m-1)(3n-2) \\ + & (3m-1)(3n-1) \end{array}$

which we can rewrite as the product

$\left(3m-2+3m-1\right)\left(3n-2+3n-1\right) = (6m-3)(6n-3)=(2m-1)(2n-1)9,$

which is exactly what we wanted.

We now move on to prove the result for any positive integer $k$ . We want to show that the sum of numbers in the square that’s $nth$ from the left and $mth$ from the top is equal to

$(2n-1)(2m-1)T_{k-1}^2.$

Let’s call the square in question $T.$

Using similar reasoning as above, you can see that the numbers in the first row of $T$ are

$\left(k(m-1)+1\right)$

times the numbers in the first row of $S$ . This means that the sum of the first row of numbers in $T$ is

$\left(k(m-1)+1\right)\left(2(n-1)T_{k-1}\right).$

Similarly, the sum of the second row of numbers in $T$ is

$\left(k(m-1)+2\right)\left(2(n-1)T_{k-1}\right).$

We can continue in this vein until we come to the sum of the last row of numbers in $T,$ which is

$\left(k(m-1)+(k-1)\right)\left(2(n-1)T_{k-1}\right).$

Adding up these sums of the rows of $T$ gives

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