Coefficients of $1,x,x^2$ in $((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$ 
Finding coefficients of $x^0,x^1,x^2$ in
$((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
where there are $k$ parenthesis in the left side

Try:
Let $P(x)=((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Let we assume
$P(x)=a_{k}+b_{k}x+c_{k}x^2+d_{k}x^3+\cdots\cdots $
For constant term put $x=0$, we have
$$((\cdots (0-2)^2-2)^2-2)^2\cdots )-2)^2=a_{k}$$
Now i did not know how to find coeff. of $1,x,x^2$ in $P(x)$ . Thanks
 A: Let the $x^2,x,1$ coefficients of the polynomial formed from the expression with $k$ brackets be $a_k,b_k,c_k$ respectively. We have
$$a_1,b_1,c_1=1,-4,4$$
$$a_{k+1},b_{k+1},c_{k+1}=2a_k(c_k-2)+b_k^2,2b_k(c_k-2),(c_k-2)^2$$
We first notice that $c_k=4$ for all $k\ge1$, so we reduce the recurrences left to solve to
$$a_{k+1},b_{k+1}=4a_k+b_k^2,4b_k$$
Solving the recurrence for $b_k$ gives $b_k=-4^k$ and
$$a_{k+1}=4a_k+4^{2k}$$
the solution of which is OEIS A166984:
$$a_k=\frac{16^k-4^k}{12}$$
Thus the three lowest-degree terms of the expression with $k$ are
$$\frac{16^k-4^k}{12}x^2-4^kx+4$$

The $x^3$ coefficient
In the same vein, denote the $x^3$ coefficient for $k$ brackets $d_k$, so
$$d_1=0\qquad d_{k+1}=2(d_k(c_k-2)+a_kb_k)=4d_k-\frac{16^k-4^k}6\cdot4^k$$
$$=4d_k-\frac{64^k-16^k}6$$
This may be solved in a similar manner to $a_k$ to obtain
$$d_k=\frac{-64^k+5\cdot16^k-4\cdot4^k}{360}$$
A: Lets consider the cases $k=1,2,3,4$:$$P_1(x)=(x-2)^2=x^2-4x+4,\\ P_2(x)=(P_1(x)-2)^2=P_1(x)^2-2P_1(x)+4=x^4-8x^3+20x^2-16x+4,\\ P_3(x)=(P_2(x)-2)^2=x^8+\dots+336x^2-64x+4,\\ P_4(x)=x^{16}+\dots+5440x^2-256x+4.$$
From this it is reasonable to conjecture that the coefficient of $x$ in $P_k(x)$ is $4^k$. Moreover, a quick check using OEIS yields the conjecture that the coefficient of $x^2$ in $P_k(X)$ is given by the sequence $$a_1=1,\\ a_2=20,\\ a_n=20a_{n-1}+64a_{n-2}.$$ This sequence can be written in a compact formula $$a_n=\frac{4\cdot 16^{n+1}-4^{n+1}}{3}.$$
You can prove both conjectures by induction on $k$.
I will show you how to do for the coefficient of $x$. The (harder) coefficient of $x^2$ I leave to you.
We proved the conjecture for $k=1,2,3,4$. This sets up induction. Now assume we know it up to $P_k(x)$. Consider $$P_{k+1}(x)=(P_k(x)-2)^2=P_k(x)^2-4P_k(x)+4.$$ Write $P_k(x)=\sum\limits_{i=0}^{2^k}c_ix^i$. The terms with $x$ in the expression above are $$c_0c_1x+c_1c_0x-4c_1x=(2c_0c_1-4c_1)x.$$ Since $c_0=4$, this yields $8c_1-4c_1=4c_1$. Now use the induction hypothesis to deduce that the coefficient equals $4\cdot 4^k=4^{k+1}$.
