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?
 A: 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 $A(n)$.
A: 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.
A: This is not a proof by induction, but is included for information of those who might be interested. 
Note that
$$r^2=\binom r2+\binom {r+1}2$$
Hence
$$\begin{align}
&\color{blue}{1^2}+\color{purple}{3^2}+\color{green}{5^2}+\cdots+\color{red}{(2n-1)^2}\\
&=\color{blue}{\binom 12+\binom 22}+
\color{purple}{\binom 32+\binom 42}+
\color{green}{\binom 52+\binom 62}+\cdots +
\color{red}{\binom {2n-1}2+\binom {2n}2}\\
&=\sum_{r=1}^{2n}\binom r2\\
&=\binom {2n+1}3
=\frac {n(2n-1)(2n+1)}3\qquad\blacksquare\end{align}$$
