Class Review: Pentagonal Numbers

Class confused me today, but my favorite part had to be when we talked about pentagonal numbers. I didn’t even know those were a thing. Basically you start with 1 (just go with it). The next pentagonal number comes from drawing a pentagon, so you get five points, therefore 5 is the next pentagonal number. For the next number, you extend that first pentagon, and draw another bigger pentagon around it, using three dots as the base instead of only two. The next number of dots you get is 12! Then 22, 35, and so on! One of the equations for pentagonal numbers is [n(3n-1)]/2.

The FOMy connection would be that, once you have the equation for pentagonal numbers, you can prove it by induction. Induction is a wonderfully simple proof device, as long as your already have the answer. That is, proof by induction revolves around the fact that you already have a claim to prove. It does not help you to explore the actual problem to figure out a claim.

But in this case, we have an equation!!

First off, we’d start with the base case, we’ll use one. If we plug 1 into the equation, we get one! Meaning that the first pentagonal number is 1. So the base case of one works.

Next is the induction step. We’ll assume that [k(3k-1)]/2 is true. Now we just need to connect the k case to the [{k+1}(3{k+1}-1)]/2 case.

Does anyone have any ideas on this? The usual method would be to take out something from the k+1 equation, so that you have a k case, which was can assume that part is true, and then usually the base case or something similar, but I can’t think of how to pull this equation apart.

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