I was in class the other day (no, really!) and I was doodling Pascal's triangle. (The lecture was on recurrence relations; I wasn't too far afield.)

I know that the terms of Pascal's triangle are the coefficients of a binomial expansion. For example, if I wished to expand (a+b)³

I could read the third row in the triangle (rows are indexed from zero, which represents the coefficient to the (a+b)⁰ term). I can see that I get {1, 3, 3, 1} which is interpreted as 1a³ + 3a²b + 3ab² + b³.

I was toying with the trinomial expansion, of the form (a + b + c)ⁿ. Expanding with n equal to 2 leads us to:

a² + 2ab + b² + 2bc + c² + 2ac
This is tacky. The 'ab' term is directly between the a² and b² terms, which makes sense. The 'ac' term, though, really should be between the a² and c² terms to make this picture symmetric, which I knew it should be.

Expanding with n = 3 makes the situation even more apparent, or more obfuscated, depending on whether you're already doodling triangles on your notepad.

a³ + 3a²b + 3ab² + 6abc +3a²c + 3ac² + b³ + 3b²c + 3bc² + c³

The terms involving a and c want even more to be 'between' the a³ and c³ terms. 3bc² feels like it should be closer to c³ than to b³, and 3ac² seems like it should be equidistant but closer to a rather than b. Further, the 6abc term seems to want to be directly between a³, b³, and c³.

So, I drew them in as a triangle.

This is already looking better - the symmetry seems much more reasonable.

Now, if we take the n=1 and n=2 triangles and overlay them with the n=3 triangle, we can see that the coefficients can be derived in a process analagous to the flat Pascal's triangle - by adding the above adjacent terms.