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?
 A: The RHS should be $$1^{ 2 }+3^{ 2 }+...+\left( 2n-1 \right) ^{ 2 }+\left( 2n+1 \right) ^{ 2 }=\underset { \left( n\left( 2n-1 \right) \left( 2n+1 \right)  \right) /3 }{ \underbrace { 1^{ 2 }+3^{ 2 }+...+(2n-1)^{ 2 } }  } +{ \left( 2n+1 \right)  }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right)  }{ 3 } +{ \left( 2n+1 \right)  }^{ 2 }=\\ =\frac { n\left( 2n-1 \right) \left( 2n+1 \right) +3{ \left( 2n+1 \right)  }^{ 2 } }{ 3 } =\frac { \left( 2n+1 \right) \left( 2{ n }^{ 2 }-n+6n+3 \right)  }{ 3 } =\frac { \left( 2n+1 \right) \left( 2{ n }^{ 2 }+5n+3 \right)  }{ 3 } =\frac { \left( n+1 \right) \left( 2n+1 \right) \left( 2n+3 \right)  }{ 3 } $$
A: \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*}
A: For $n=1$ the formula holds,

$$S_n=1^2+3^2+5^2+\cdots+(2n-1)^2$$
  $$S_{n+1}=1^2+3^2+\cdots+n^2+(2n+1)^2$$
  $$\bbox[5px,border:2px solid red]{S_{n+1}-S_n=(2n+1)^2}$$

Now if,
$$S_n=\frac{n(2n-1)(2n+1)}{3}$$$$$$
$$S_{n+1}=\frac{(n+1)(2n+1)(2n+3)}{3}$$$$$$
$$S_{n+1}-S_n=\frac{(n+1)(2n+1)(2n+3)}{3}-\frac{n(2n-1)(2n+1)}{3}$$$$$$
$$=(2n+1)\frac{(n+1)(2n+3)-n(2n-1)}{3}$$$$$$
$$=(2n+1)\frac{(n)(2n+3)-n(2n-1)+(2n+3)}{3}$$$$$$
$$=(2n+1)\frac{6n+3}{3}$$$$$$
$$S_{n+1}-S_n=(2n+1)^2$$$$$$
Hence proved....
A: For $2n-1 = 1$:
$$1^2 = \frac{1(2*1-1)(2*1+1)}{3} = \frac{3}{3} = 1$$
Assuming the formula works for $n=2k-1$, where $k$ is some integer, for $n=2(k+1)-1$:
$$1^2+3^2+...+(2k-1)^2+(2(k+1)-1)^2 =$$
$$ \frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 \stackrel{?}{=} \frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} $$
By expanding the right hand side of the equation:
$$\frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} = \frac{4k^3+12k^2+11k+3}{3}$$
And by simplifying the left hand side of the equation:
$$\frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 = \frac{4k^3-k}{3} + (2k+1)^2 = \frac{4k^3-k}{3}+4k^2+4k+1 = \frac{4k^3+12k^2+11k+3}{3}$$
Finally:
$$ \frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 = \frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} $$
And so for any $n$, where $2n-1 \in \mathbb{N}$, this formula holds:
$$1^2 + 3^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
A: Here's a non-induction approach, provided for information only:
$$\begin{align}
&1^2+3^2+5^2+\cdots+(2n-1)^2\\
&=\sum_{r=1}^n(2r-1)^2\\
&=\sum_{r=1}^n\binom {2r-1}2+\binom {2r}2\\
&=\sum_{s=1}^{2n}\binom s2\\
&=\binom {2n+1}3
&&=\frac {(2n+1)(2n)(2n-1)}{1\cdot 2\cdot 3}\\
& &&=\frac 13n(2n-1)(2n+1)\\
& && =\frac 13 n(4n^2-1)\end{align}$$
