Try to factor first the that we need, the sum at n+1 is
Use PMI to prove for all natural numbers n,
summation (i=1 to n) (2i-1)^3=n^2((2n^2)-1)
So that statement is P(n). Clearly I can check P(1) is true. No problem there.
P(n+1) is summation (i=1 to n+1)(2i-1)^3=summation (i=1 to n)(2i-1)^3+(2(n+1)-1)^3=n^2((2n^2)-1)+(2(n+1)-1)^3
I expand those and get 2n^4+8n^3+11n^2+6n+1
Now I know I need to show this is equal to (n+1)^2((2(n+1)^2)-1). I know (n^2+2n+1)(2n^2+4n+1) is the same thing as this. Bascially I just need help filling in between....rusty algebra I suppose.