Finding closed form of sequence $0, 2, 10, 28, 60, 110, 182,...$

Question

Need to find closed form of this sequence.

$0, 2, 10, 28, 60, 110, 182, 280, 408, 570, 770,...$

I've tried some obvious ideas but yet to have a luck.

Answer 1

The sequence of first differences is $$ 2,8,18,32,50,72,\ldots $$ which is double the sequence of squares. Hence the $n$th term of the original sequence is twice the sum of the first $n-1$ squares, or $$ \frac13 (n - 1)n(2n - 1). $$

Answer 2

Also you can try this approach. I did not work it out myself, but a way to check correctness is to compare to other answers.

Given your sequence:

\begin{array}{|c|c|c|c|c|c|c|c|} \hline n= & 1 & 2 & 3 & 4 & 5 & 6 & 7 & ... \\ \hline f(n)= & 0 & 2 & 10 & 28 & 60 & 110 & 182 & ...\\ \hline \end{array}

Notice that the third differences of the sequence are constant. So the closed form is a cubic polynomial $$ f(n) = an^3 + bn^2 + cn + d $$

So we plug in the first 4 values of $n$ to get 4 equations with 4 unknowns. Then solve these linearly for: $$\begin{cases} 0 = a+b+c+d \\ 2 = a(2^3) + b(2^2) + c(2)+d \\ 10 = a(3^3) + b(3^2) + c(3)+d \\ 28 = a(4^3) + b(4^2) + c(4)+d \end{cases}$$