<s>[INST] Sum of a Sequence of Odd Numbers that are Squared What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + \ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.
This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):


I am a little stuck on what to do next and how to obtain $\frac{n (4n^2 - 1)}{3}$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)
 [/INST] Your approach is almost correct. Check again your steps. At the end you should have
$$\begin{align}(k+\mathbf{2})^3-1
&=6(1+3^2+\dots+k^2)+12(1+3+\dots+k)+\underbrace{(8+8+\dots+8)}_{\text{$(k+1)/2$ times}}\\
&=6(1+3^2+\dots+k^2)+12\left(\frac{k+1}{2}\right)^2+8\mathbf{\left(\frac{k+1}{2}\right)}.\end{align}$$
Hence 
$$6(1+3^2+\dots+k^2)=(k+\mathbf{2})^3-1-12\left(\frac{k+1}{2}\right)^2-8\mathbf{\left(\frac{k+1}{2}\right)}$$
and it follows that
$$\sum_{j=1}^n(2j-1)^2=1+3^2+\dots+k^2=\frac{k(k+2)(k+1)}{6}=\frac{n (4n^2 - 1)}{3}$$
where $n=(k+1)/2$.
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