Casey gave us the example today of creating an induction proof for the series of pentagonal numbers, without first knowing the formula that we are trying to prove.
The nth pentagonal number in the number of distinct dots in a pattern of dots consisting of outlines of regular pentagons with sides of length n dots, when the pentagons are overlaid so that they share a single vertex. Thought it would be difficult to build the formula on our own, there is a known formula for the pentagonal number series:
pn = (3n2-n)/2 for n ≥ 1.
Here is a proof by induction for this formula.
Base case for n = 1
When n = 1, the visual representation of this formula is a single dot. So pn should equal 1.
pn = (3(1)2-1)/2 = (3-1)/2 = 1.
Inductive step
We now want to show that this is true for pk and pk+1. Assume that pk is true. That is pk = (3k2-k)/2. We want to show that pk+1 = (3(k+1)2-(k+1))/2 is also true.
We also know, that to get to the next k, we must add 3k+1 dots. (If you are not convinced, look at the first few pentagonal numbers.)
pk+1 = (3(k+1)2-(k+1))/2
pk+1 = (3(k2+2k+1)-k-1)/2
pk+1 = (3k2+6k+3-k-1)/2
pk+1 = ((3k2-k)+6k+2)/2
pk+1 = pk+(6k+2)/2
pk+1 = pk+3k+1.
inFOMation 