Question

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

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

coeff. of $x^{2} \Rightarrow 2 + \ ^{n}C_{2}$

$$2 + \frac{n(n - 1)}{2} $$Coeff. of $x^{3} = 2.\ ^{n}C_{1} + \ ^{n}C_{3}$

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

Now according to question

$${n + 2n + \frac{n(n - 1)(n - 2)}{6} = 2\left\lbrack 2 + \frac{n(n - 1)}{2} \right\rbrack }{3n + \frac{n(n - 1)(n - 2)}{6} = 4 + n(n - 1) }{\Rightarrow n^{3} - 9n^{2} + 26n - 24 = 0 }{\Rightarrow n = 2,3,4\ \Rightarrow n = 3,4 }$$Now checking for $n = 2$

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

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

Required sum of values of ' n '

$$= 2 + 3 + 4 = 9 $$