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?

The RHS should be $$1^{ 2 }+3^{ 2 }+...+\left( 2n-1 \right) ^{ 2 }+\left( 2n+1 \right) ^{ 2 }=\underset { \left( n\left( 2n-1 \right) \left( 2n+1 \right)  \right) /3 }{ \underbrace { 1^{ 2 }+3^{ 2 }+...+(2n-1)^{ 2 } }  } +{ \left( 2n+1 \right)  }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right)  }{ 3 } +{ \left( 2n+1 \right)  }^{ 2 }=\\ =\frac { n\left( 2n-1 \right) \left( 2n+1 \right) +3{ \left( 2n+1 \right)  }^{ 2 } }{ 3 } =\frac { \left( 2n+1 \right) \left( 2{ n }^{ 2 }-n+6n+3 \right)  }{ 3 } =\frac { \left( 2n+1 \right) \left( 2{ n }^{ 2 }+5n+3 \right)  }{ 3 } =\frac { \left( n+1 \right) \left( 2n+1 \right) \left( 2n+3 \right)  }{ 3 } $$
