<s>[INST] 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?
 [/INST] Here's a non-induction approach, provided for information only:
$$\begin{align}
&1^2+3^2+5^2+\cdots+(2n-1)^2\\
&=\sum_{r=1}^n(2r-1)^2\\
&=\sum_{r=1}^n\binom {2r-1}2+\binom {2r}2\\
&=\sum_{s=1}^{2n}\binom s2\\
&=\binom {2n+1}3
&&=\frac {(2n+1)(2n)(2n-1)}{1\cdot 2\cdot 3}\\
& &&=\frac 13n(2n-1)(2n+1)\\
& && =\frac 13 n(4n^2-1)\end{align}$$
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