*Admin note: This is my first experiment getting latex formatting working for formulas so there's going to be more sigma notation than I would use in class. We're also on a break until next week.*

A triangle number is a sum 1 + 2 + ... + n which is also written $\sum\limits_{i=1}^n i$.

Two cool theorems that can be fairly easily proved are:

- $\sum\limits_{i=1}^n i$ = $\frac{n(n+1)}{2}$

- $(\sum\limits_{i=1}^n i)^2 = \sum\limits_{i=1}^n i^3$

There's a geometric interpretation that makes this fairly obvious:

**New: cool triple triangle proof http://t.co/2QwN1sKdeY**

However, I don't see the kids discovering this on their own but I suppose I could provide cut outs and have them do the transforms manually.

The inductive proof on the other hand might be more easily discovered but I foresee two problems.

1. Most of the kids don't have a good grasp on multiplying expressions like $(a + b)^2 = a^2 + 2ab + b^2$. We could review this fact again (which we did once during the pythagorean theorem session.

2. Secondly I don't think most kids have done any inductive proofs. So it would probably also require a session just on that to work out the principles using simpler problems.

**TODO:**what would a good set look like?

With those techniques in hand the following would be workable:

Let $s_1 = (\sum\limits_{i=1}^n i)$

Then

$(\sum\limits_{i=1}^{n+1} i)^2 = (s_1 + (n+1))^2$

Which expands to:

$s_1^2 + 2*s_1*(n+1) + (n+1)^2$

Then since we also know $s_1 = \frac{n(n+1)}{2}$ we can substitute that into the second term.

$s_1^2 + 2*\frac{n(n+1)}{2}*(n+1) + (n+1)^2$

which simplifies to

$s_1^2 + n(n+1)^2 + (n+1)^2$

And then using the distribute law to factor out n+1 you get:

$s_1^2 + (n + 1) (n+1)^2 = s_1^2 + (n+1)^3 $

That's enough to use the inductive principle i.e. each successive term means adding the next cube and I think I could move the kids through steps via a worksheet type approach.

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