algebra precalculus - The sum of $1^2+7^2+13^2+cdots+n^2,$ where $n =1 mod6$

Find the sum of this progression in the terms of $n$

$$1^2+7^2+13^2+\cdots+n^2,$$ where $n =1 \mod6$.

Are Bernoulli's numbers involved in this? Please help.

Answer

So, we need $\displaystyle\sum_{r=0}^m(6r+1)^2$

which is
$$36\sum_{r=0}^mr^2+12\sum_{r=0}^mr+\sum_{r=0}^m1$$

$$=36\sum_{r=1}^mr^2+12\sum_{r=1}^mr+(m+1)$$

References :

Proving the sum of the first $n$ natural numbers by induction

How to get to the formula for the sum of squares of first n numbers?