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?
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