Seriously cool! On the assumption your DH hasn't already explained it to DS, here's an attempt at an explanation that might make sense to your DS (I'm making the assumption that he spotted the pattern, which is impressive, rather than proving it always holds, which would be mind-blowing). It'll be morally the same as what your DH did with triangle numbers, of course, but perhaps more approachable.

Write the numbers out in rows:

1 2 3 4 (row 1)
6 8 10 12 (row 2)
15 18 21 24 (row 3)
etc., so that the row number is the number we're skip-counting by in that row.


What we're going to do is to show that the pattern continues for ever. We can see the pattern holds in row 1, and what we're going to do is to show that if it holds in any row, it holds in the next row as well. So row 1's pattern will force row 2's pattern to hold, and row 2's pattern will force row 3's pattern to hold, and so on as far as you like. (You may, but need not, tell him that this is called induction.)

But what exactly is it that holds in row 1? What matters is not just that 4 is divisible by 2. Actually, 4 is 2 x 1 (this row's number) x 2 (the next row's number). Let's say it like this: the last number in row 1 is the number of unit cubes that are in a cuboid with height 2, depth 1 and width 2. The pattern we're going to show holds is that the last number in any row, let's call it the nth row, is the number of unit cubes in a cuboid with height 2, depth n (this row's number), and width n+1 (the next row's number).

It might be a good idea at this point to build such a cuboid, say for n=3, e.g. in Lego.

Now, how much do we have to add to get from the last number in row n to the last number in row n+1? In row n+1 we're doing skip counting by n+1, so we have to add n+1, four times.

Imagine taking four n+1-long bits of Lego (actually do it, and make them a different colour from the old cuboid, perhaps). Put them in a two by two by n+1 cuboid shape, and line it up with the width of the cuboid we built for the last number on row n, so that the two cuboids are stuck together on the 2 by n+1 face.

Look, we get a new cuboid, and the number of unit cubes in this must be the last number in the n+1th row, because we got it by doing a new row of skip counting starting from the last number in the nth row.

What are the dimensions of this cuboid? It's still got height 2, and it's still got width n+1, but now its depth is its old depth, which was n, plus the extra bit we just added, which has depth 2. So it's 2 by n+1 by n+2.

That proves that the last number in the n+1th row is divisible by n+2, so the pattern holds for another row.

Turn the cuboid round so that the n+2 length side (which is the depth at the moment) is the width, and we're ready to do exactly the same thing again for the next row... and we could keep doing this for ever!

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