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

Question

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.

The Question has meant $$n\equiv1\pmod6$$ right? – lab bhattacharjee Jul 18 '14 at 16:54 — lab bhattacharjee, Jul 18 '14 at 16:54

score 11 · Accepted Answer · edited Apr 13 '17 at 12:20

11

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 :

edited Apr 13 '17 at 12:20

Community

answered Jul 18 '14 at 16:46

lab bhattacharjee

score 0 · Answer 2 · answered Jul 18 '14 at 17:54

0

After simplifications :

$$ \sum_{k=0}^{n} (6*k+1)^2 = (n+1)\left(12n^2+12n+1\right) $$

Shadock

answered Jul 18 '14 at 17:54

ParaH2

Note that the usage of $n$ here is not the same as in the problem statement. – Semiclassical Jul 18 '14 at 18:23
And why ? He write the sum with n^2 and if n=1 mod 6 so n=6k+1 with k in Z so i don't understand you :/ – ParaH2 Jul 18 '14 at 18:52
...because I'm being silly. Woops. – Semiclassical Jul 18 '14 at 19:06
He writes* Woops ^^ – ParaH2 Jul 18 '14 at 19:25

2 Answers2