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?

Let your statment be $A(n)$. You want to show it holds for all $n \in \mathbb{N}$. You use the principle of induction to establish a chain of implications starting at $A(1)$ (you did that one).
What is left to show is
$$
A(n) \Rightarrow A(n+1)
$$
This means you consider $n$ fixed and try to proof $A(n+1)$. For this you can assume that $A(n)$ is true, or in another variant of the principle of induction: $A(1), \dotsc, A(n)$ are true.
You have:
\begin{align}
A(n) &: \sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3} \\
A(n+1) &: \sum_{k=1}^{n+1} (2k-1)^2 = \frac{(n+1)(2(n+1)-1)(2(n+1)+1)}{3} \\
\end{align}
One way is to split
$$
\sum_{k=1}^{n+1} (2k-1)^2 = \left(\sum_{k=1}^{n} (2k-1)^2 \right) + (2(n+1)-1)^2
$$
and to use $A(n)$.
