Debbie's fascination that the sum of the first n odd numbers is n2 is understandable, but this is also easy to prove. The key is that the "triangle sum", the sum of the numbers from 1 to n, is equal to n(n+1)/2. (See note at bottom for a quick visual proof.) The triangle sum is easy to demonstrate by induction; it is often used in a classroom setting as an introductory example of how induction works. In fact it's proved in a writeup in the induction node. Hey wait a minute, it's proved in Debbie's writeup in the induction node.
So Debbie--you already know a way to demonstrate this.
n2 = 1 + 3 + 5 + 7 + ... + 2n-1
= 1+2+3+4+5+6+7+ ... + 2n-1 (sum of all integers up to 2n-1)
- 2 + 4 + 6 + ... +2n (less the even integers up to 2n)
But we can represent both of those as a triangle sum T: the first is just T2n-1, the sum of the numbers from 1 to (2n-1), and the second, the sum of the even integers up to 2n is just 2Tn, twice the sum of the integers up to n:
= T2n-1 - 2*Tn-1
= --------- - 2*(--------)
= (2n-1)*n - (n-1)*n
= n*n = n2
Pretty neat, but now you need something else to ponder. How about the sum of the first n integers squared (1 + 4 + 9 + ...)? That's trickier, but doable. See the first couple chapters of Concrete Mathematics by Knuth et al if you get stuck; they do a great job with it.
(NOTE: Tn is called the "triangle sum" because it can be visualized as a triangle. Imagine a set of bowling pins arranged as:
o o o
o o o o
o o o o o
The number of pins is clearly 1 + 2 + ... + n, where n is the last row. Now put that triangle together with another triangle arranged upside-down, and you get an n x (n+1) rectangular matrix of pins. So the number of pins in the triangle is half that, n*(n+1)/2.)