Just to make a few comments:

1) As noted, If we have a list of values $a_0,a_1,\cdots, a_n$ of integers then there is a (unique) polynomial $f(t)$ of degree no more than $n$ with integer coefficients which maps integers to integers and such that $f(k)=k$ for $0 \le k \le n$.

2) There is a method involving the differences and differences of the differences etc. which reveals $f(n)$ as an integer linear combination of the polynomials $\binom{t}{j}$ for $0 \le j \le n$. Furthermore, the polynomials of this form are exactly the polynomials sending integers to integers. These (specialized) Newton Polynomials are very similar to the Taylor series which uses the basis $\frac{t^k}{k!}$

3) If you just want the next term (as predicted by this polynomial) then you don't need to explicitly find the polynomial, just extend the differences. Many test takers realize this.
$$\begin{matrix}2&\ &6&\ &12&\ &20&\ &\mathbf{30}\\\ &4&&6&&8&&\mathbf{10}&\\\ &&2&&2&&\mathbf{2}&&\end{matrix}$$ corresponds to $f(n)=2+4n+2\binom{n}{2}=n^2+3n+2$

4) There is also a polynomial of degree 3 that gives $2,6,12,20,\mathbf{2011}$ so there is no unique extension.

5) If the given sequence is $1,2,4,8,?$ then the polynomial interpolation gives $15$ next from $\binom n0+\binom n1+\binom n2+\binom n3$ although most tests would favor another continuation.