5.29 Simplify the expression
$$
\left(\left(\cdots\left(\left((x-2)^{2}-2\right)^{2}-2\right)^{2}-\cdots-2\right)^{2}-2\right)^{2} \text {, }
$$

where there are $k$ nested parentheses, and combine like terms to obtain a polynomial. Find the coefficient of $x^{2}$ in this polynomial.
[Solution] Let the obtained polynomial be $P_{k}(x)$, and set
$$
P_{k}(x)=A_{k}+B_{k} x+C_{k} x^{2}+\cdots \text {. }
$$

Since
$$
\begin{aligned}
P_{k}(0) & =\left(\left(\cdots\left(\left((0-2)^{2}-2\right)^{2}-2\right)^{2}-\cdots-2\right)^{2}-2\right)^{2} \\
& =4,
\end{aligned}
$$

we have
$$
P_{k}(x)=4+B_{k} x+C_{k} x^{2}+\cdots
$$

Next, we use mathematical induction to prove:
$$
B_{k}=-4^{k}, C_{k}=4^{k-1} \cdot \frac{4^{k}-1}{4-1} .
$$

In fact, when $k=1$,
$$
P_{1}(x)=(x-2)^{2}=x^{2}-4 x+4 \text {, }
$$

thus
$$
B_{1}=-4, C_{1}=1 \text {. }
$$

Assume that for $k=m$, we have
$$
B_{m}=-4^{m}, C_{m}=4^{m-1} \cdot \frac{4^{m}-1}{4-1} .
$$

Then
$$
\begin{aligned}
P_{m+1}(x) & =\left(\left(4+B_{m} x+C_{m} x^{2}+\cdots\right)-2\right)^{2} \\
& =\left(2+B_{m} x+C_{m} x^{2}+\cdots\right)^{2} \\
& =4+4 B_{m} x+\left(B_{m}^{2}+4 C_{m}\right) x^{2}+\cdots
\end{aligned}
$$

Therefore
$$
\begin{aligned}
B_{m+1} & =4 B_{m}=4 \cdot\left(-4^{m}\right)=-4^{m+1} \\
C_{m+1} & =\left(-4^{m}\right)^{2}+4 \cdot\left(4^{m-1} \cdot \frac{4^{m}-1}{4-1}\right) \\
& =4^{m} \cdot \frac{4^{m+1}-1}{4-1} .
\end{aligned}
$$

By the principle of mathematical induction, for any natural number $k$, we have
$$
C_{k}=4^{k-1} \cdot \frac{4^{k}-1}{4-1} .
$$

The coefficient of $x^{2}$ is thus $C_{k}$.