$(...((x-2)^2-2)^2-...-2)^2\;\;\text{(n parentheses)}$

Question

Find the coefficient next to $x^2$ in the development of a polynomial: $$(...((x-2)^2-2)^2-...-2)^2\;\;\text{(n parentheses)}$$ I am not sure but I think we get the coefficient in the following way:

We have something like this quasi-recursive formula:

$$...+k_1\cdot x^2+k_2\cdot x+2\;\text{and}\; k_3=k_2^2+4k_1$$ $$k=2^{2(n+1)}+4\cdot k_2$$

$$(...((x-2)^2-2)^2-...-2)^2$$ $$(x^2-4x+4-2)^2=(x^2-4x+2)^2$$ $$((x^2-4x+2)^2-2)^2=(x^4+16x^2+4-8x^3+4x^2-16x-2)^2=(x^4-8x^3+20x^2-16x+2)^2$$ $$((x^4-8x^3+20x^2-16x+2)^2-2)^2$$

Hopefully:

$$20=4^2+4\;\text{next}\;16^2+4\cdot 4=336$$

Answer 1

We have that

$$(x-2)^2=x^2-4x+4$$

$$((x-2)^2-2)^2=(x^2-4x+2)^2=\ldots+20x^2-16x+4$$

$$(((x-2)^2-2)^2-2)^2=(\ldots-16x+2)^2=\ldots 336x^2-64x+4$$

therefore we can guess that

• coefficient for $x$ are $-4,-16,-64,\ldots \implies k_2(n)=-(4^n)$
• coefficient for $x^2$ are $1,20,336,\ldots \implies k_1(n)=(k_2(n-1))^2+4k_1(n-1)$

which can be proved rigoursly by induction, therefore we have

$$k_1(n)=4k_1(n-1)+4^{2(n-1)}$$

which leads to

$$k_1(n)=\frac13\left(4^{2n-1}-4^{n-1}\right)$$