<s>[INST] Find $2^2 + 5^2 + 8^2 + ... + (3n-1)^2 $ 
First prove that $1^2 + 2^2 + 3^2 +...+ n^2 = \frac{n(n+1)(n+2)}{6}$, then find $$2^2 + 5^2 + 8^2 + ... + (3n-1)^2.$$

I can prove the first part but I have no idea about the second part.
 [/INST] Since
$$\sum_{k=1}^n (3n-1)^2 = 9\sum_{k=1}^n k^2-6\sum_{k=1}^n k + \sum_{k=1}^n 1$$
and
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$, use your previous result to get
$$\begin{align}
\sum_{k=1}^n (3n-1)^2 &= 9\sum_{k=1}^n k^2-6\sum_{k=1}^n k + \sum_{k=1}^n 1 \\
&=9\left(\frac{n(n+1)(n+2)}{6}\right)-6\left( \frac{n(n+1)}{2}\right)+n \\
\end{align}$$ and work from there
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