Example 2-46 Suppose there are $n$ elliptic curves, each pair of which intersects at two points, and no three elliptic curves intersect at a single point. How many parts does such $n$ ellipses divide the plane into?
An ellipse (I) divides the plane into inner and outer parts. Two ellipses divide the plane into 4 parts, the boundary of the second ellipse (II) is divided into two parts by (I), which are precisely the boundaries of the newly added regions. Similarly, the third ellipse (III) is divided into 4 parts by the first two ellipses, thus dividing the plane into \(4 + 4 = 8\) parts, as shown in Figure 2-13. If \(n-1\) ellipses divide the plane into \(a_{n-1}\) regions, the \(n\)-th ellipse intersects each of the previous \(n-1\) ellipses at two points, resulting in \(2(n-1)\) intersection points, and thus is divided into \(2(n-1)\) segments, each of which is the boundary of a new region. Therefore, the number of new regions added is \(2(n-1)\), and if \(n\) ellipses divide the plane into \(a_{n}\) parts, then:
\[
\begin{array}{l}
a_{n}=a_{n-1}+2(n-1) \\
a_{1}=2
\end{array}
\]

Let
\[
\begin{array}{c}
G(x)=a_{1}+a_{2} x+a_{3} x^{2}+a_{4} x^{3}+\cdots \\
x: \quad a_{2}=a_{1}+2 \\
x^{2}: \quad a_{3}=a_{2}+2 \cdot 2 \\
x^{3}: \quad a_{4}=a_{3}+2 \cdot 3 \\
+\quad \quad \cdots \\
\hline G(x)-2=x G(x)+2 x+2 \cdot 2 x^{2}+2 \cdot 3 x^{3}+\cdots
\end{array}
\]

Thus \((1-x) G(x)=2+2 x+2 \cdot 2 x^{2}+2 \cdot 3 x^{3}+\cdots\)
\[
=2\left(1+x+2 x^{2}+3 x^{3}+\cdots\right)
\]

Therefore,
\[
\begin{aligned}
G(x) & =2\left(1+x+2 x^{2}+3 x^{3}+\cdots\right) /(1-x) \\
& =2\left(1+x+2 x^{2}+3 x^{3}+\cdots\right)\left(1+x+x^{2}+\cdots\right) \\
& =2\left(1+2 x+4 x^{2}+7 x^{3}+11 x^{4}+16 x^{5}+22 x^{6}+29 x^{7}+\cdots\right)
\end{aligned}
\]

Thus,
\[
a_{1}=2, a_{2}=4, a_{3}=8, a_{4}=14, a_{5}=22, a_{6}=32, a_{7}=44, \cdots
\]

In general, \(a_{n}=2\left(1+\sum_{k=1}^{n-1} k\right)=2\left(1+\frac{n(n-1)}{2}\right)=n(n-1)+2\)
Another method:
\[
a_{n}=a_{n-1}+2 n-2
\]

Let \(\alpha\) be a particular solution to the non-homogeneous recurrence relation, and let
\[
\alpha=k_{0}+k_{1} n+k_{2} n^{2}
\]

Substituting into equation (2-4), we get
\[
\begin{array}{c}
k_{0}+k_{1} n+k_{2} n^{2}=k_{0}+k_{1}(n-1)+k_{2}(n-1)^{2}+2 n-2 \\
2 k_{2} n+k_{1}-k_{2}=2 n-2 \\
k_{2}=1, k_{1}-k_{2}=-2 \\
k_{1}=-2+k_{2}=-2+1=-1 \\
\alpha=n^{2}-n=n(n-1)
\end{array}
\]

Thus,
\[
a_{n}=a_{n-1}
\]

The general solution to the homogeneous recurrence relation is \(a_{n}=c\), where \(c\) is a constant.
The solution to the non-homogeneous recurrence relation (2-4) is the sum of the particular solution \(\alpha\) and the general solution to the homogeneous recurrence relation, i.e.,
\[
a_{n}=n(n-1)+c
\]

Using the initial condition \(a_{1}=2\), we get \(c=2\), so the solution to the problem is
\[
a_{n}=n(n-1)+2
\]