General term of sequence 3

Question

Let be $$6,8,12,15,20,24,30,35,42,… $$ a sequence of natural numbers. Guessing the recurrence then using generating functions I can prove that general term of sequence is $$a_n=\frac{1}{8}(2n^2+20n+45+3(-1)^n)$$ I think that my proof is too long. Any shorter more elementary proof.

Answer 1

Let $n=1$, the right side of your equation is $(1+20+45-3)/8=63/8$, so something's wrong.

Taking the odd index terms $6,12,20,30,42,\dots$, the first differences are $6,8,10,12,\dots$, the second differences are constant $2,2,2,\dots$, so the sequence is quadratic, $n^2+3n+2$.

The even index terms $8,15,24,35,\dots$, first difference $7,9,11,\dots$, second difference $2,2,\dots$, also quadratic, $n^2+4n+3$.

So we get $a_n=((n+1)/2)^2+3(n+1)/2+2$ if $n$ is odd, $a_n=(n/2)^2+2n+3$ if $n$ is even. Can you see how to combine those into a single formula?