Let $f_{0}(x)=|x|, \quad f_{1}(x)=\left|f_{0}(x)-1\right|, \quad f_{2}(x)=\left|f_{1}(x)-2\right|, \quad \ldots f_{n}(x)=\left|f_{n-1}(x)-n\right|$. The graph of the function $f_{n}(x)$ intersects the $x$-axis at exactly two points. What is the area of the region enclosed by this function curve and the $x$-axis?
Let the sought area be $t_{n}$. For the values $n=0,1,2$, plotting the function $f_{n}$ (Figure 1), we can see that $t_{0}=0$, $t_{1}=1$, $t_{2}=7$, and with the notation $a_{n}=n(n+1) / 2$, the following statement is true for $n=0,1,2$:

$$
\begin{array}{ll}
\text { if } & x \leq-a_{n}, \quad \text { then } \quad f_{n}(x)=-x-a_{n} \\
\text { if } & -a_{n}<x<a_{n}, \quad \text { then } \quad 0<f_{n}(x) \leq n \\
\text { if } & a_{n} \leq x, \quad \text { then } \quad f_{n}(x)=x-a_{n}
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_21549b0c1c77f0523e06g-1.jpg?height=726&width=1584&top_left_y=480&top_left_x=256)

Since $a_{n}-a_{n-1}=n$ and $f_{n}(x)=\left|f_{n-1}(x)-n\right|$, if statement (1) is true for $(n-1)$, then it is also true for $n$ (Figure 2). Therefore, by the principle of mathematical induction, (1) holds for all $n$. Based on Figure 2, the area $t_{n}$ is obtained by subtracting the area of the region bounded by the curve of the function $f_{n}$ between points $E$ and $F$ and the segment $EF$ from the area of the trapezoid $A D F E$. This is nothing more than the reflection of the curve of $f_{n-1}$ between points $B$ and $C$ over the line $y=n / 2$, so the area to be subtracted is exactly $t_{n-1}$. The area of the trapezoid is

$$
B E \frac{A D+E F}{2}=n\left(a_{n}+a_{n-1}\right)=n^{3}
$$

thus $t_{n}=n^{3}-t_{n-1}$. Writing this relationship for $n$ replaced by $(n-1),(n-2), \ldots$ and adding them with alternating signs, we get that for even $n$,

$$
t_{n}=n^{3}-(n-1)^{3}+(n-2)^{3}+\ldots+2^{3}-1^{3}
$$

and for odd $n$,

$$
t_{n}=n^{3}-(n-1)^{3}+(n-2)^{3}+\ldots-2^{3}+1^{3}
$$

since $t_{0}=0$. This solves the problem.

Remark. For $t_{n}$, not only the expressions in (2) and (3), but also a closed form can be given. For this, we start from the following identity:

$$
8 n^{3}=\left[4 n^{3}+6 n^{2}-1\right]+\left[4(n-1)^{3}+6(n-1)^{2}-1\right]
$$

Comparing this with the relationship $8 n^{3}=8 t_{n}+8 t_{n-1}$, we get that the sum of consecutive terms of the sequence $8 t_{n}-\left[4 n^{3}+6 n^{2}-1\right](n=0,1,2, \ldots)$ is 0, and since the value of the expression $8 t_{n}-4 n^{3}+6 n^{2}-1$ is +1 for $n=0$, the terms of the sequence alternate between +1 and -1. From this, we get that $8 t_{n}-\left[4 n^{3}+6 n^{2}-1\right]=(-1)^{n}$, so

$$
t_{n}=\frac{4 n^{3}+6 n^{2}-1+(-1)^{n}}{8}
$$