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Let $\{f_n(x)\}$ be a sequence of polynomials defined inductively as

$$ f_1(x)=(x-2)^2$$

$$f_{n+1}(x) = (f_n(x)-2)^2, \: \: \: n \geq 1$$

Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then

  1. $a_n=4, \: b_n=-4^n$
  2. $a_n=4, \: b_n=-4n^2$
  3. $a_n=4^{(n-1)!}, \: b_n=-4^n$
  4. $a_n=4^{(n-1)!}, \: b_n=-4n^2$
in Combinatory by Veteran (425k points)
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1 Answer

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Answer: $\mathbf A$

Explanation:

$\mathrm {f_1(x) = (x-2)^2\;;\;f_{n+1} = (f_n-2)^2} \;\;\; \text {[Given]}$ 

$\mathrm {f_1(x) = (x-2)^2 = 4 - 4 x +x^2}$

$\mathrm {f_2(x) = (f_1(x)-2)^2 = (2-4x+x^2)^2 = 2^2 + 2.2(-4x) +\cdots=4-4^2x+\cdots}$

$\mathrm {f_3(x) = (f_2(x)-2)^2 = (4-2^4x+\cdots-2)^2 = (2-2^4x+\cdots)^2\cdots =2^2-2.2.4^2x+\cdots=2^2-4^3x+\cdots}$

$\mathrm {f_4(x) = (f_3(x)-2)^2=2^2-4^4x+\cdots}$

$\mathrm {f_5(x) = (f_4(x)-2)^2 = 2^2-4^5x+\cdots}$

So, $\mathrm {a_n = 4\;,\;b_n = -4^{(n^{th}\;term \;\;of\; 1,2,3,4,5)}}$

$\therefore \mathbf A $ is the correct option.

by Boss (14.1k points)
edited by

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