Aspire Faculty ID #17994 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

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

Solution

$(x^4 + 2x^2 + 1)\left({}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + \cdots\right)$

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 $= 2 + 3 + 4 = 9$

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