Question: Let the function $f_{0}(x)=|x|, f_{1}(x)=\left|f_{0}(x)-1\right|$, $f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$
Let $S_{n}$ denote the corresponding area, then from Figure 1, it is easy to see:
$$
\begin{array}{l}
S_{0}=0, S_{1}=1, \\
S_{2}=7 .
\end{array}
$$

Let $f_{0}(x)=|x|$, 
$$
f_{n}(x)=\left|f_{n-1}(x)-n\right|
$$
$(n \in N)$, and $S_{n}$ represents the area of the closed figure formed by the graph of $f_{n}(x)$ and the $x$-axis, then $S_{n}=$ ?

First, let's examine the properties of $f_{n}(x)$. By definition and mathematical induction, it is easy to see that $f_{n}(x)$ is an even function. Therefore, its graph is a right-angled broken line symmetric about the $y$-axis, and it passes through the point $\left(0,\left[\frac{n+1}{2}\right]\right)$. For example:
- The breakpoints of $f_{3}(x)$ are $( \pm 1,1),( \pm 3,3),( \pm 6,0)$,
- The breakpoints of $f_{4}(x)$ are $( \pm 1,3),( \pm 3,1),( \pm 6,4)$, $( \pm 10,0)$.

Thus, $\frac{S_{3}}{2}=\frac{1}{2}(2+1)(1-0)+\frac{1}{2}(1+3)(3-1)$
$$
\begin{aligned}
& +\frac{1}{2} \times 3 \times(6-3), \\
S_{3}= & 20 \\
\frac{S_{4}}{2}= & \frac{1}{2}(2+3)(1-0)+\frac{1}{2}(3+1)(3-1) \\
& +\frac{1}{2}(1+4)(6-3)+\frac{1}{2} \times 4 \times(10-6), \\
S_{4}= & 44 .
\end{aligned}
$$

It is not difficult to generalize that when $n=2 m-1(m \in N)$,
$$
\begin{aligned}
S_{2 m-1}= & (1+3+\cdots+(2 m-1))(2 m-1) \\
& +(2+4+\cdots+(2 m-2)) \cdot 2 m \\
= & (4 m-3) m^{2}, \\
S_{i m}= & (1+3+\cdots+(2 m-1))(2 n+1)+ \\
& (2+4+\cdots+2 m) \cdot 2 m \\
= & (4 m+3) m^{2} .
\end{aligned}
$$

Thus, $S_{n}=\frac{1}{4}\left(2 n^{3}+3 n^{2}-\frac{1+(-1)^{n+1}}{2}\right)$, $n=0,1,2, \cdots$.
(This can be proven by mathematical induction)