<s>[INST] Is this proof by induction for a sum of odd squares correct? Statement: $1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n/3)*(2n-1)*(2n+1)$
Proof by induction
-Base case: when $n = 1$
$1^2 = 1/3 * (2 * 1 -1) * (2 * 1 +1) = 1$
$1=1$ hence statement holds for $n = 1$
-Inductive step
assume $n = k$ is true
then $1^2 + 3^2 + 5^2 + .. + (2k-1)^2 = (k/3) * (2k - 1) * (2k + 1)$
let $n = k+1$
then $1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$
it follows that $1^2 + 3^2 +5^2 .. + (2k-1)^2 = (k/3) * (2k-1) * (2k+1)$
such that $(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2 = ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$
hence: $(4k^3+12k^2 + 11k + 3)/3  = (4k^3+12k^2 + 11k + 3)/3$
thus the statement holds for $n+1$ and the statement is therefore true
Is this proof correct? Am I missing anything or detail that I can add?
 [/INST] Your line "hence something complicated is equal to itself" is unconvincing and suggests that you might do better with a different order, perhaps like $$1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$$ $$=(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2$$ $$=(4k^3+12k^2 + 11k + 3)/3$$ $$= ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$$ $$= (n/3)*(2n-1)*(2n+1) \text{ when } n=k+1$$
</s>