Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$1^2+3^2+...+(2n-1)^2+(2n+1)^2=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$n(2n-1)(2n+1)/3+3(2n+1)^2/3=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$4n(2n-1)(2n+1)/3=(n+1)(2n+3)/3$
What is wrong?

\begin{eqnarray*} 
1^2+3^2+5^2+\cdots+(2n-1)^2=\frac{(2n-1)n(2n+1)}{3}
\end{eqnarray*}
So
\begin{eqnarray*} 
1^2+3^2+5^2+\cdots+(2n-1)^2 + (2n+1)^2 &=& \frac{(2n-1)n(2n+1)}{3} + (2n+1)^2 \\
&=& \frac{(2n+1)}{3} \underbrace{\left(  n(2n-1) +3(2n+1) \right)}_{2n^2+5n+3} \\
&=&  \frac{(2n+1)(n+1)(2n+3)}{3}.
\end{eqnarray*}
