2. There are $n$ circles on a plane, each pair of which intersects at two points, and no three circles have a common point. They divide the plane into $f(n)$ regions, such as $f(1)=2, f(2)=4$, etc. The expression for $f(n)$ is ( ).
A. $2^{n}$
B. $n^{2}-n+2$
C. $2^{n}-(n-1)(n-2)(n-3)$
D. $n^{3}-5 n^{2}+10 n-4$
2. B.

Since the $n$-th circle intersects with the previous $n-1$ circles at $2(n-1)$ points, it is divided into $2(n-1)$ arcs, thus the number of plane regions increases by $2(n-1)$. Therefore, $f(n)-f(n-1)=2(n-1)$. Hence, $f(n)-f(1)=2[1+2+\cdots+n-1]=$ $n^{2}-n$, which means $f(n)=n^{2}-n+2$.