What is the coefficient and constant term in the following sequence defined recursively? Let $f_n(x)$ be a sequence of polynomials defined inductively as
$f_1(x) = (x - 2)^2$
$f_{n+1}(x) = (f_n(x) - 2)^2$ $; n \ge 1$
Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then
$(A) a_n = 4, b_n =  -4^n (B) a_n = 4, b_n =  -4n^2
(C) a_n = 4^{(n-1)!} , b_n =  -4^n 
(D) a_n = 4^{(n-1)!}, b_n =  -4n^2.$
By simpllifying the expression, we can derive the term, ie, by expanding out the powers with $(a+b)^2$ formula recursively. But is there an analytical approach to solve this problem?
 A: A straightforward approach is to see how $a_n$ and $b_n$ change when $n$ is increased by $1$.
The constant term of $f_1(x)$ is $4$. Suppose that $f_n(x)$ for some $n$; then $f_n(x)=xg_n(x)+4$ for some polynomial $g_n(x)$ (why?), so
$$f_{n+1}(x)=\left(\big(xg_n(x)+4\big)-2\right)^2=\big(xg_n(x)+2\big)^2=x^2\big(g_n(x)\big)^2+4xg_n(x)+4\;.$$
The first two terms of $x^2\big(g_n(x)\big)^2+4xg_n(x)+4$ are multiples of $x$, so the constant term of $f_{n+1}(x)$ must be $4$. Thus, $a_n=4$ for each $n\in\Bbb Z^+$. (Technically this is a proof by mathematical induction.)
Now let’s look at the $x$ term. We can do pretty much the same thing: we know that $$f_n=x^2h_n(x)+b_nx+4$$ for some polynomial $h_n(x)$ (why?), so
$$\begin{align*}
f_{n+1}(x)&=\big(x^2h_n(x)+b_nx+2\big)^2\\
&=\Big(x\big(xh_n(x)+b_n\big)+2\Big)\\
&=x^2\big(xh_n(x)+b_n\big)^2+4x\big(xh_n(x)+b_n\big)+4\\
&=x^2\Big(x\big(xh_n(x)+b_n\big)^2+4h_n(x)\Big)+4b_nx+4\;.
\end{align*}$$
The first of those three terms is a multiple of $x^2$, so $4b_nx$ is the $x$ term of $f_{n+1}(x)$, and we see that $b_{n+1}=4b_n$. Thus, $b_2=4b_1$, $b_3=4b_2=4^2b_1$, $b_4=4b_3=4^3b_1$, and in general to get from $b_1$ to $b_n$ we must multiply by $4$ a total of $n-1$ times: 
$$b_n=4^{n-1}b_1=4^{n-1}(-4)=-4^n\;.$$
