Generating a formula for an Octagonal figurative number

Question

 Generate a formula for the nth Octagonal number by relating to one or more figurate numbers with fewer than seven sides(by relating to hexagonal, pentagonal, square, oblong or triangular numbers).

Answer 1

This is going to be hard to do without being able to write actual fractions, but bear with me.  I'll assume that you've been given the formulae for pentagonal and hexagonal numbers, and we have square numbers. 

 

So here we go:

 

square           4 sides     formula is n^2

pentagonal     5 sides     formula is (1/2)n(3n-1)

hexagonal      6 sides     formula is n(2n-1)

 

These all look different but the trick is to force them to look the same. Now I don't know which formulae you started with.  There are many variations that come out to the same thing, so if yours aren't mine, try to work out what I'm doing and why so you can do the same thing to yours.  This is the crux move, the thing that makes it work, because it is the only way to know if there is a pattern to these very different looking formulae.

 

I'll start with the hexagonal formula and try to make it look like the pentagonal one.  This means I need a 1/2, but if I put one in I have to also multiply by 2 to keep things the same:

     n(2n-1) = (1/2)(2)n(2n-1)

Now I distribute the 2 into the second part and get (1/2)n(4n-2).  This is what I'm hoping for because now the pentagonal and hexagonal look like this:

 

pentagonal     5 sides     formula is (1/2)n(3n-1)
hexagonal      6 sides     formula is (1/2)n(4n-2)

 

So I got from one to the next by adding one to the 3 and subtracting 1 from the -1.  If this pattern continues I'll get to octagonal in two steps from hexagonal, and that means add 2 and subtract 2, so the formula would be (1/2)n(6n-4)

 

But I want to check first by going backwards and seeing if I get the right formula for square numbers.  So start from pentagonal and subtract 1 from the 1st term and add 1 to the second, so (1/2)n(3n-1) becomes (1/2)n(2n-0) which simplifies to n(n) = n^2, which is right.  If I do it again, (1/2)n(2n-0) becomes (1/2)n(n+1), which is the formula for triangular numbers, so I'm happy now.

 

The formula for octagonal should then be (1/2)n(6n-4).  I can distribute the 1/2 to get (1/2)n(3n-2)

 

 

 

 

 

 

 

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