Series: 3, 6, 11, ... continues ambiguously, so let's assume that the gap between the terms is growing by 2; but the difference between perfect squares of the natural numbers also grows by 2.
If we subtract 2 from each term we get 1, 4, 9, ... and therefore an=n2+2.
The common difference is (n+1)2+2-(n2+2)=n2+2n+1-n2=2n+1.
The sum Sn of n terms = ∑(i2+2) for 1≤i≤n where i is an integer.
Sn=2n+∑i2=2n+⅙n(n+1)(2n+1).
S1=2+⅙(2)(3)=3; S2=4+⅙(2)(3)(5)=9=3+6; S3=6+⅙(3)(4)(7)=20=3+6+11, ...
∑i2=⅙n(n+1)(2n+1) can be fairly easily proved, but, unless you need the proof, I've omitted it here. I'll provide it if you need it---just make a comment on my answer.