Proof by induction that $1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$ I need to know if I am doing this right. I have to prove that 
$1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$
So first I did the base case which would be $1$.
$1^2 = (1(2(1)-1)(2(1)+1)) / 3
1 = 3/3
1 = 1$ Which is right.
Then I assumed true for k so
$1^2 + 3^2 + 5^2 + ... + ((2k-1)^2 = k(2k-1)(2k+1)) / 3$
This is where I get lost. I know that the next step is to prove its true for k+1, but I am lost on what to do here. Do I actually simplify down what I assumed true for k? Am I right so far?

Hypothesis: $$\sum \limits_{k=1}^n (2k-1)^2 = \dfrac{n(2n-1)(2n+1)}{3} $$
For $n+1$: $$\sum \limits_{k=1}^{n+1} (2k-1)^2 = \dfrac{(n+1)(2n+1)(2n+3)}{3} = \dfrac{4n^3 - n}{3} + (2n+1)^2 = \dfrac{n(2n-1)(2n+1)}{3} + (2n+1)^2$$
which is exactly what we wanted.
