"Baby" Proof by Induction (AKA--Ends up being I need help with algebra lol)

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.

Try to factor first the $\displaystyle (n+1)^2$ that we need, the sum at n+1 is
$\displaystyle \begin{array}{ccc} S(n+1) & = & n^2(2n^2-1)+(2n+1)^3 \\ & = & n^2(2n^2-1)+(n+1)^3 + 3(n+1)^2n + 3(n+1)n^2 + n^3 \\ & = & n^2(2n^2-1+3(n+1)+n) +(n+1)^3 + 3(n+1)^2n \\ & = & 2n^2(n+1)^2 + (n+1)^3 + 3(n+1)^2n \\ & = & (n+1)^2(2n^2 + 4n + 1) \\ & = & (n+1)^2(2(n+1)^2-1) \end{array} $