Example 2. Given n spheres, each pair of which intersects in a circle. How many regions do these n spheres divide the space into?

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Example 2. Given n spheres, each pair of which intersects in a circle. How many regions do these n spheres divide the space into?
Consider the following problem first:
On a sphere, there are $\mathrm{n}$ circles, and every two circles have two common points. Suppose such $\mathrm{k}$ circles divide the sphere into $\mathrm{f}(\mathrm{k})$ regions, then $(\mathrm{k}+1)$ such circles divide the sphere into $\mathrm{f}(\mathrm{k}+1)$ regions. The $(k+1)$-th circle intersects with the previous $\mathrm{k}$ circles at $2 \mathrm{k}$ points, and is divided into $2 \mathrm{k}$ arcs, and each small arc divides its region into two. Therefore, we have the recursive relation:
$$
\mathrm{f}(\mathrm{k}+1)-\mathrm{f}(\mathrm{k})=2 \mathrm{k} \text {. }
$$

For this formula, let $k=1,2,3, \cdots,(n-1)$
we get:
$$
\left.\begin{array}{l}
f(2)-f(1)=2 \times 1, \\
f(3)-f(2)=2 \times 2, \\
f(4)-f(3)=2 \times 3, \\
\ldots \ldots \\
f(n)-f(n-1)=2 \times(n-1)
\end{array}\right\} \text { equations }
$$
$\left\{\begin{array}{l}\text { a total of }(n-1) \\ \text { equations }\end{array}\right.$
Adding all of them, we simplify to:
$$
\begin{array}{l}
f(n)-f(1)=2[1+2+3 \cdots \cdots+(n-1)] \\
=n(n-1) . \\
\because f(1)=2, \quad \therefore f(n)=n^{2}-n+2 .
\end{array}
$$

To answer the question:
For $\mathrm{k}$ mutually intersecting spheres, they divide the space into $F(\mathrm{k})$ regions. We know that the $(k+1)$-th sphere intersects with the previous $\mathrm{k}$ spheres at $\mathrm{k}$ circles, and these circles divide the $(\mathrm{k}+1)$-th sphere into $\left(\mathrm{k}^{2}-\mathrm{k}+2\right)$ regions, and each small region divides its space region into two. Therefore, we have the recursive relation: $F(k+1)-F(k)=k^{2}$ $-\mathrm{k}+2$. For this formula, let $\mathrm{k}=1,2,3, \cdots$, $(n-1)$, we get $(n-1)$ equations, add them all, and simplify.
$$
\begin{aligned}
& \because F(1)=2, \\
& \therefore F(n)=F(1)+\left[1^{2}+2^{2}+3^{2}+\cdots\right. \\
+ & \left.(n-1)^{2}\right]-[1+2+3+\cdots+(n-1)] \\
+ & 2(n-1) \\
= & \frac{(n-1) n(2 n-1)}{6}-\frac{(n-1) n}{2}+2 n \\
= & \frac{n\left(n^{2}-3 n+8\right)}{3} .
\end{aligned}
$$

These $\mathrm{n}$ spheres divide the space into $\frac{1}{3} n\left(n^{2}-3 n+8\right)$ regions.
(Author's affiliation: Gaoyou County High School, Jiangsu Province)