Induction!

This is inspired by a Bluesky post from Laurence Tribe¹.

https://bsky.app/profile/tribelaw.bsky.social/post/3lepw4xn6h22x

The years Y = 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025 … all follow the pattern Y = square of (1 + … + n) = 1 cubed + … + n cubed.

That’s a pattern I hadn’t seen before. Proving it looks like a job for mathematical induction. What’s that, you ask? Well, there are several forms, but the one that applies here is:

1. Start with a logical expression with one variable (refer to it as F(i))
2. Show that F(1) is true.
3. Show that, given F(i), F(i+1) is true.
4. Now F(i) is true for all positive whole numbers.

Why? Well, in (2) we showed F(1) is true. By (3) that means F(2) is true. And that means (again by (3)) F(3) is true. And you can keep applying (3) forever.

Consider the sum of all numbers 1 to N ². We express that via sigma notation as ∑_j=0ⁱj
(read as “the sum from j equals zero to i of j”.)³. That equals i(i+1)/2. Proof?

2) For i=1, 1(2)/2 does equal 1.
3) Assume i(i+1)/2 = ∑_j=0ⁱj . Now consider (i+1)(i+2)/2. This is i(i+1)/2 + 2(i+1)/2 = i(i+1)/2 + i+1
This is ∑_j=0ⁱi + i+1 or ∑_j=0ⁱ⁺¹j.

Now let’s get back to Mr. Tribe’s pattern⁴, which can be written as
(∑_j=0ⁱj)² = ∑_j=0ⁱj³
and see where we get.

2) For i=1, 1² does equal 1³.
3) Assume (∑_j=0ⁱj)² = ∑_j=0ⁱj³ . This time, consider (∑_j=0ⁱ⁺¹j)². This expands to (∑_j=0ⁱj)² + 2(i+1)(∑_j=0ⁱj) + (i+1)². But (∑_j=0ⁱj) is i(i+1)/2 (we just proved this) so this is (∑_j=0ⁱj)² + 2i(i+1)²/2 + (i+1)² or (∑_j=0ⁱj)² + (i+1)(i+1)² or (∑_j=0ⁱj)² + (i+1)³. This is ∑_j=0ⁱj³ + (i+1)³ or ∑_j=0ⁱ⁺¹j³.

That’s what we needed to prove, so indeed (∑_j=0ⁱj)² = ∑_j=0ⁱj³ for all positive whole numbers.

1. Yes, that Laurence Tribe.
2. As an example, for N=3, this would be 1 + 2 + 3, which equals 6.
3. That’s not formatted quite right,but it’s as close to correct as I can get with this kind of blog post.
4. As a simple example, (1+2+3+4)² is 100, as is 1+8+27+64.