This is inspired by a Bluesky post from Laurence Tribe1Yes, that Laurence Tribe..
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.So the last pre-2025 year of this form was 1296 = (1 + 2 + … + 8) squared = 36 squared = 1 cubed + … + 8 cubed.The first post-2025 will be 3025 = 55 squared.
— Laurence H. Tribe (@tribelaw.bsky.social) 2025-01-02T01:24:26.053Z
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:
- Start with a logical expression with one variable (refer to it as F(i))
- Show that F(1) is true.
- Show that, given F(i), F(i+1) is true.
- 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 2As an example, for N=3, this would be 1 + 2 + 3, which equals 6.. We express that via sigma notation as ∑j=0ij
(read as “the sum from j equals zero to i of j”.)3That’s not formatted quite right,but it’s as close to correct as I can get with this kind of blog post.. 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 pattern4As a simple example, (1+2+3+4)2 is 100, as is 1+8+27+64., 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.
Come for the snark, stay for the math.
I remember the first time I banged my head against mathematical induction. After a while, it just became automatic.
It’s like a recursive program. You don’t seem to have done enough, but it works.