296 *. Determine the coefficient of $x^{2}$ after expanding the brackets and combining like terms in the expression

$$
\underbrace{\left(\ldots\left(\left((x-2)^{2}-2\right)^{n}-2\right)^{2}-\ldots-2\right)^{2}}_{k \text { times }}
$$
296. First, let's find the free term, which will result from the expression

$$
\underbrace{\left.\left(\ldots(x-2)^{2}-2\right)^{2}-\ldots-2\right)^{2}}_{k}
$$

if we expand the brackets and combine like terms. It is equal to the value of this expression at \( x=0 \), i.e.,

$$
\begin{aligned}
& \underbrace{\left(\ldots\left(\left((-2)^{2}-2\right)^{2}-2\right)^{2}-\ldots-2\right)^{2}}_{k}= \\
& =\underbrace{\left(\ldots\left((4-2)^{2}-2\right)^{2}-\ldots-2\right)^{2}}_{k-1}=
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-316.jpg?height=85&width=562&top_left_y=339&top_left_x=286)

$$
\begin{aligned}
& \ldots=\left((4-2)^{2}-2\right)^{2}=(4-2)^{2}=4
\end{aligned}
$$

Now, let's denote \( A_{k} \) as the coefficient of \( x \), \( B_{k} \) as the coefficient of \( x^{2} \), and \( P_{k} x^{3} \) as the sum of terms that contain \( x \) in higher powers. Then we have:

$$
\left(\ldots\left((x-2)^{2}-2\right)^{2}-\ldots-2\right)^{2}=P_{k} x^{3}+B_{k} x^{2}+A_{k} x+4
$$

On the other hand,

$$
\begin{gathered}
\left(\ldots\left(\left((x-2)^{2}-2\right)^{2}-2\right)^{2}-\ldots-2\right)^{2}= \\
k \text { times } \\
=\left(\frac{\left(\left(\ldots\left((x-2)^{2}-2\right)^{2} \ldots-2\right)^{2}-2\right)^{2}=}{k-1}\right)^{2} \\
=\left[\left(P_{k-1} x^{3}+B_{k-1} x^{2}+A_{k-1} x+4\right)-2\right]^{2}= \\
=\left(P_{k-1} x^{3}+B_{k-1} x^{2}+A_{k-1} x+2\right)^{2}= \\
=P_{k-1}^{2} x^{6}+2 P_{k-1} B_{k-1} x^{5}+\left(2 P_{k-1} A_{k-1}+B_{k-1}^{2}\right) x^{4}+ \\
+\left(4 P_{k-1}+2 B_{k-1} A_{k-1}\right) x^{3}+\left(4 B_{k-1}+A_{k-1}^{2}\right) x^{2}+ \\
+4 A_{k-1} x+4=\left[P_{k-1}^{2} x^{3}+2 P_{k-1} B_{k-1} x^{2}+\right. \\
\left.+\left(2 P_{k-1} A_{k-1}+B_{k-1}^{2}\right) x+\left(4 P_{k-1}+2 B_{k-1} A_{k-1}\right)\right] x^{3}+ \\
\cdot \quad+\left(4 B_{k-1}+A_{k-1}^{2}\right) x^{2}+4 A_{k-1} x+4
\end{gathered}
$$

From this, we have

$$
A_{k}=4 A_{k-1}, B_{k}=A_{k-1}^{2}+4 B_{k-1}
$$

Since \((x-2)^{2}=x^{2}-4 x+4\), then \(A_{1}=-4\). Therefore, \(A_{2}=-4 \cdot 4=-4^{2}, A_{3}=-4^{3}, \ldots\), and generally \(A_{k}=-4^{k}\).

Now, let's compute \(B_{k}:\)

$$
\begin{gathered}
B_{k}=A_{k-1}^{2}+4 B_{k-1}=A_{k-1}^{2}+4\left(A_{k-2}^{2}+4 B_{k-2}\right) \Rightarrow \\
=A_{k-1}^{2}+4 A_{k-2}^{2}+4^{2}\left(A_{k-3}^{2}+4 B_{k-3}\right)= \\
=A_{k-1}^{2}+4 A_{k-2}^{2}+4^{2} A_{k-3}^{2}+4^{3}\left(A_{k-4}^{2}+4 B_{k-4}\right)=\ldots \\
\cdots=A_{k-1}^{2}+4 A_{k-2}^{2}+4^{2} A_{k-3}^{2}+\ldots \\
\cdots+4^{k-3} A_{2}^{2}+4^{k-2} A_{1}^{2}+4^{k-1} B_{1}
\end{gathered}
$$

Substituting here

$$
\begin{aligned}
& B_{1}=1, A_{1}=-4, A_{2}=-4^{2} \\
& A_{3}=-4^{3}, \ldots, A_{k-1}=-4^{k-1}
\end{aligned}
$$

we get:

$$
\begin{gathered}
B_{k}=4^{2 k-2}+4 \cdot 4^{2 k-4}+4^{2} \cdot 4^{2 k-6}+\ldots+4^{k-2} \cdot 4^{2}+4^{k-1} \cdot 1= \\
=4^{2 k-2}+4^{2 k-3}+4^{2 k-4}+\ldots+4^{k+1}+4^{k}+4^{k-1}= \\
=4^{k-1}\left(1+4+4^{2}+4^{3}+\ldots+4^{k-2}+4^{k-1}\right)= \\
=4^{k-1} \frac{4^{k}-1}{4-1}=\frac{4^{2 k-1}-4^{k-1}}{3}
\end{gathered}
$$