Here is the promised geometric visualization of the fact that the $n$th square number is the sum of the first $n$ odd numbers. I hope it may be a bit more enjoyable for readers, who I know are mostly Olly well-wishers rather than math enthusiasts.
I actually developed this visualization between the first and second proofs I shared yesterday. As I knew the proof by induction would probably practically write itself, I got it out of the way first. I then moved on to look at the problem geometrically, hoping to gain insight that I could use to write a more illuminating proof. This geometric approach turned out to be more closely related to the proof by induction than to the second proof I came up with later, though.
Claim: The $n$th square number is the sum of the first $n$ odd numbers.
First of all, how would we represent a square number visually? One way is as a collection of dots arranged in a square, as shown below.
Second, how can we represent an odd number visually? Well, every odd number is one more than an even number, and any even number, being divisible by 2, can be represented as two rows of the same number of dots. Thus, an odd number can be represented as two rows of the same number of dots, plus one more. (Except for 1, which is simply 1 dot.)
Now imagine rearranging each odd number into an L shape as shown below. These L-shaped figures also have two rows of the same number of dots, plus one more.
Next, notice how the series of L-shaped figures representing the odd numbers nest together to form a square.
Thus, the claim holds at least up to $n=5$. To draw a square representing the next square number, you would add a row and a column to this square, which you can see would be the same as adding another L-shaped figure representing the next largest odd number. So the claim must hold for each higher $n$ as well.
I also played around with a similar visualization using even numbers. The L-shaped figures representing odd numbers can be adapted to instead represent even numbers by removing the dot in the corner. When you do that, the nested figures look like this:
The center diagonal is missing from each square. Since it contains a number of dots equal to the side length of the square, the number of dots left in the $n$th square is $n^2-n$.
Accounting for the fact that the first even number, 2, corresponds to the second square, this leads to the formula
$$\sum_{i=1}^n2i=(n+1)^2-(n+1)=(n+1)[(n+1)-1]=(n+1)(n)\text{.}$$
This is what you would expect given the well known formula for the sum of the first $n$ natural numbers and can also be proven by induction. I also checked the sum of this formula and the one from yesterday. The sum of the first $n$ odd natural numbers plus the sum of the first $n$ even natural numbers should equal the sum of the first $2n$ natural numbers, which it does.
$$n^2+(n+1)(n)=n[n+(n+1)]=n(2n+1)=\frac{2n(2n+1)}{2}$$
Your next-to-last formula really amounts to saying that sum(2i) = 2 sum(i), doesn’t it? See where 2 sum(i) lives in the picture? All this is very cool! And is there a way to post formulas in the comments?
Yes, now that you mention it, I can see the two triangles in each of my squares. Nifty.
The plug-in that allows me to insert math in my posts covers the comments, too. You enclose the code in single dollar signs for in-line math and double dollar signs for math centered on its own line. The math uses LaTeX codes.
\$\sum_{i=1}^n2i\$ gives: $\sum_{i=1}^n2i$
\$\$\sum_{i=1}^n2i\$\$ gives:
$$\sum_{i=1}^n2i$$
Ah, so
$$
\sum_{i=0}^n 2i=2\sum_{i=0}^n i=2\frac{n(n+1)}{2},
$$
then, does it?
My goodness, that’s annoying to type on a phone!
It must be. It’s a bit annoying on a full keyboard. LaTeX is a source of nostalgia for me, though. I can’t remember if I ever used it in your classes, but I used it for most of my homework in Anand’s analysis class because writing made my hands hurt.