Check the forward differences of the sequence.
If [tex]\{a_n\} = \{2,6,12,20,30,42,\ldots\}[/tex], then let [tex]\{b_n\}[/tex] be the sequence of first-order differences of [tex]\{a_n\}[/tex]. That is, for n ≥ 1,
[tex]b_n = a_{n+1} - a_n[/tex]
so that [tex]\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}[/tex].
Let [tex]\{c_n\}[/tex] be the sequence of differences of [tex]\{b_n\}[/tex],
[tex]c_n = b_{n+1} - b_n[/tex]
and we see that this is a constant sequence, [tex]\{c_n\} = \{2, 2, 2, 2, \ldots\}[/tex]. In other words, [tex]\{b_n\}[/tex] is an arithmetic sequence with common difference between terms of 2. That is,
[tex]2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2[/tex]
and we can solve for [tex]b_n[/tex] in terms of [tex]b_1=4[/tex]:
[tex]b_{n+1} = b_n + 2[/tex]
[tex]b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2[/tex]
[tex]b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2[/tex]
and so on down to
[tex]b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)[/tex]
We solve for [tex]a_n[/tex] in the same way.
[tex]2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)[/tex]
Then
[tex]a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)[/tex]
[tex]a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))[/tex]
[tex]a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))[/tex]
and so on down to
[tex]a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2[/tex]
[tex]\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}[/tex]
