Summing Products

Some days your mind wanders into an interesting puzzle: not necessarily because it’s a difficult puzzle, but because it has familiar result. Then the puzzle becomes, how are the two things linked?

For example, I had cause to add up all of the numbers in the times tables - let’s say the grid from $1\times1$ to $10\times10$.

I did the natural thing for a mathematician: I split it into ten arithmetic sequences - the numbers from $1\times 1$ to $10 \times 1$ sum to 55; the numbers from $1 \times 2$ to $10 \times 2$ sum to $55 \times 2$, and so on.

Then I have $55 \times (1 + 2 + \dots + 10) = 55^2 = 3025$.

In general, the times table grid from $1 \times 1$ to $n \times n$ gives a sum of $\left(\frac{n(n+1)}{2}\right)^2$, or $\frac{n^2(n+1)^2}{4}$.

But hang on a minute

That’s the sum of the cubes from 1 to $n$! Why on earth should that be? I don’t see any cubes!

After a lot of thought, I found them.

Let’s just look at the products1 that have a 10 in them - from $1 \times 10$ to $10 \times 10$ down to $10 \times 1$, nineteen products in all.

Imagine each of them as a rectangle. We have a $10 \times 10$ square. We also have $9 \times 10$ and $10 \times 1$, which we can stick together to make a square. Similarly, $8 \times 10$ and $10 \times 2$. The eighteen non-square products can be paired up to make nine ten-by-ten squares - added to the $10 \times 10$, that gives us a $10 \times 10 \times 10$ cube.

And we’re left with the times tables up to $9\times 9$, on which we can play the same trick. Each ‘band’ of the times table grid can be folded together to make a cube!

I thought that was a lovely, unexpected connection. I’d be thrilled to hear if you had another way of explaining it - the comments are open!


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. in our original 10 by 10 grid, although it works generally []


2 comments on “Summing Products

  • Benjamin Leis

    I was recently looking at the same problem by coincidence in a slightly different form

    3 5
    7 9 11
    13 15 17 19

    Note each row sums to a cube and you can bring in the triangle numbers with a little manipulation.

    The algebra for the induction of wjy this works isn’t too bad but it rang some bells about the well known identity that the square of a triangle nunber is the sum of cubes.

    Then I cheated because I remembered a nice geometric diagram on the wikipedia

    • Colin

      Oo! Nice diagram. And it reminds me of a related-but-different one I had in mind but didn’t ever produce, and now can’t remember how it went! There was some duplicating and folding, but I can’t find the cubes in my head any more.

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