If recursion polynomials $P_{k} (x)$ are defined as $P_{1} (x) = {(x - 2)}^{2}, P_{2} (x) = {({(x - 2)}^{2} - 2)}^{2}, P_{3} (x) = {({({(x - 2)}^{2} - 2)}^{2} - 2)}^{2} ....$ $(I n g e n e r a l P_{k} (x) = {(P_{K - 1} (x) - 2)}^{2}),$ then

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In $P_{k} (x)$ coefficient of $x^{2}$ is $\frac{4^{2 k - 1} - 4^{k - 1}}{3}$

In $P_{k} (x)$ coefficient of x is $- 4^{k}$

In $P_{k} (x)$ coefficient of x is $4^{k}$

In $P_{k} (x)$ constant term is 4

answer is A, C, D.

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Detailed Solution

In $P_{1} (x),$ constant term =4
In $P_{2} (x),$ constant term =4
In $P_{3} (x),$ constant term =4
In $P_{k} (x),$ constant term =4
$P_{1} (x) = {(x - 2)}^{2} = x^{2} + 4 - 4 x$
Coefficient of $x i n P_{1} (x) = - 4 \to (1) i s t r u e$
$P_{2} (x) = {({(x - 2)}^{2} - 2)}^{2} = {(x^{2} - 4 x + 2)}^{2} = x^{4} + 4^{2} x^{2} + 2^{2} - 8 x^{3} + 4^{2} x + 4 x^{2}$
Coefficient $x i n P_{2} (x) = - 4^{2}$
Similarly, coefficient of x in $P_{k} (x) = - 4^{k} \to (3) i s t r u e$
Now, $P_{1} (x),$ in coefficient of $x^{2} = 1 = \frac{4^{2 - 1} - 4^{1 - 1}}{3}$
In $P_{2} (x),$ Coefficient of $x^{2} = 20 = \frac{4^{2 (2) - 1} - 4^{2 - 1}}{3}$
Similarly, in $P_{k} (x),$ Coefficient of $x^{2} = \frac{4^{2 k - 1} - 4^{k - 1}}{3} \to (4) i s t r u e$