Any large square for general k

Share this page

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

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


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

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

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


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

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}   (1)

Similarly to what we did above, we can re-write this expression as

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

Using the formula for the sum of the first $k-$ integers, this becomes

  \[ \left[\frac{(k-1)k}{2}+(k-1)k(m-1)\right](2n-1)T_{k-1}, \]    

which is equal to

  \[ \frac{(k-1)k}{2}(2m-1)(2n-1)T_{k-1}. \]    

And since

  \[ \frac{(k-1)k}{2}=T_{k-1}, \]    

the sum of the numbers in $T$ is equal to

  \[ (2m-1)(2n-1)T_{k-1}^2. \]    

This is what we wanted to show.

Return to main article

  • Want facts and want them fast? Our Maths in a minute series explores key mathematical concepts in just a few words.

  • What do chocolate and mayonnaise have in common? It's maths! Find out how in this podcast featuring engineer Valerie Pinfield.

  • Is it possible to write unique music with the limited quantity of notes and chords available? We ask musician Oli Freke!

  • How can maths help to understand the Southern Ocean, a vital component of the Earth's climate system?

  • Was the mathematical modelling projecting the course of the pandemic too pessimistic, or were the projections justified? Matt Keeling tells our colleagues from SBIDER about the COVID models that fed into public policy.

  • PhD student Daniel Kreuter tells us about his work on the BloodCounts! project, which uses maths to make optimal use of the billions of blood tests performed every year around the globe.