The Maths of Christmas- Triangle Numbers Response

[PEZenfuego]

@singingbanana I was a little confused when you responded to this video. This is actually an eight minute video that Youtube apparently thought was too boring and cut down to 2 minutes. In any case, I will attempt to upload it properly so that I can actually get to my question and equation. In any case, I really appreciate your comments. They were very prompt and informative. Thank you very much.

[singingbanana]

Hi! The general, what I was doing with triangle numbers is called an arithmetic series. In most general terms, let's say you have n terms, starting from a, with a common difference d (you add d each time). Then the sum is:

a + (a+d) + (a+2d) + (a+3d) + ... + (a + (n-2)d) + (a + (n-1)d)

[singingbanana]

@singingbanana

If we add the first term and the last term we get 2a+(n-1)d.

If we add the second term and the second to last term we also get 2a+(n-1)d

If we add the third term and the third to last term we would get 2a+(n-1)d

In fact we have n/2 such pairs, so the sum is equal to (n/2)*(2a + (n-1)d)

[singingbanana]

@singingbanana That's the general equation, but I prefer to call the first term a and the last term l so the formula is

(n/2)*(a+l)

There's a lovely story (myth?) of a seven year olf Guass (famous mathematician), who was given the sum 1 + 2 + 3 + ... + 50 to keep him busy. And after a few seconds he answered 5050 because he worked out the formula.

[singingbanana]

@singingbanana This is an arithmetic sum. The next step is a geometric sum, where you add numbers that have a common ratio r, i.e. you start at a and multiply by r, so;

a + ar + ar^2 + ar^3 + .... + ar^(n-1).

Maybe I'll leave that for you to discover.