The sum of all possible values of , so that the coefficients of and in the expansion of , are in arithmetic progression, is: [2026]

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Solution

Q.
The sum of all possible values of $n \in N$ , so that the coefficients of $x, x^{2}$ and $x^{3}$ in the expansion of ${(1 + x^{2})}^{2} {(1 + x)}^{n}$ , are in arithmetic progression, is: [2026]

1 3

2 7

3 9

4 12

Ans.
(3)

$(x^{4} + 2 x^{2} + 1) ({}^{n}C_{0} x^{0} + {}^{n}C_{1} x^{1} + {}^{n}C_{2} x^{2} + {}^{n}C_{3} x^{3} + \dots)$

$Coefficient of x \Rightarrow {}^{n}C_{1}$

$Coeff. of x^{2} \Rightarrow 2 + {}^{n}C_{2} = 2 + \frac{n (n - 1)}{2}$

$Coeff. of x^{3} = 2 \cdot {}^{n}C_{1} + {}^{n}C_{3}$

$= 2 n + \frac{n (n - 1) (n - 2)}{6} (if x \geq 3)$

Now according to question

$n + 2 n + \frac{n (n - 1) (n - 2)}{6} = 2 [2 + \frac{n (n - 1)}{2}]$

$3 n + \frac{n (n - 1) (n - 2)}{6} = 4 + n (n - 1)$

$\Rightarrow n^{3} - 9 n^{2} + 26 n - 24 = 0$

$\Rightarrow n = 2, 3, 4 \Rightarrow n = 3, 4$

$Now checking for n = 2$

$\begin{matrix} Coeff. of x = 2 \\ Coeff. of x^{2} = 3 \\ Coeff. of x^{3} = 4 \end{matrix}} \Rightarrow are in A.P.$

$\Rightarrow n = 2 is also the correct choice$

$Required sum of values of n$

$= 2 + 3 + 4 = 9$

$Option (3)$

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