Q:

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.

A:

Hint
$$\sum(3k-1)^2=9\sum k^2-6\sum k+\sum 1.$$

A:

Note that
$$\sum_{k=1}^n (3n-1)^2 = 9\sum_{k=1}^n k^2-6\sum_{k=1}^n k + \sum_{k=1}^n 1$$
You know how to evaluate the first term, and you can evaluate the second term using
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$

