🎥 Video solution / Text Solution of this question is given below:
$(x^4 + 2x^2 + 1)\left({}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + \cdots\right)$
Coefficient of $x$ = ${}^nC_1$
Coefficient of $x^2$ = $2 + {}^nC_2$
Coefficient of $x^3$ = $2\cdot {}^nC_1 + {}^nC_3$
$= 2n + \frac{n(n-1)(n-2)}{6}$
Now according to question
$n + 2n + \frac{n(n-1)(n-2)}{6} = 2\left(2 + \frac{n(n-1)}{2}\right)$
$3n + \frac{n(n-1)(n-2)}{6} = 4 + n(n-1)$
$\Rightarrow n^3 - 9n^2 + 26n - 24 = 0$
$\Rightarrow n = 2, 3, 4$
Now checking for $n = 2$
Coeff of $x = 2$, coeff of $x^2 = 3$, coeff of $x^3 = 4$ ⇒ are in A.P.
Required sum $= 2 + 3 + 4 = 9$