Find the sum of the series 3+6+11+⋯,to n terms

Question

A question on sequences and series

first term-3

last term-n

common difference-+2

Answer 1

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 a_n=n²+2.

The common difference is (n+1)²+2-(n²+2)=n²+2n+1-n²=2n+1.

The sum S_n of n terms = ∑(i²+2) for 1≤i≤n where i is an integer.

S_n=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.