JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

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 :

A
$3$
B
$7$
C
$12$
D
$9$

Solution

$(1 + x^{2})^{2} = 1 + 2 x^{2} + x^{4}$ . Multiplying by $(1 + x)^{n} = \sum^{n} C_{k} x^{k}$ :

Coefficient of $x$ : $^{n} C_{1} = n$ .

Coefficient of $x^{2}$ : $^{n} C_{2} + 2 \cdot^{n} C_{0} = \frac{n ( n - 1 )}{2} + 2$ .

Coefficient of $x^{3}$ : $^{n} C_{3} + 2 \cdot^{n} C_{1} = \frac{n ( n - 1 ) ( n - 2 )}{6} + 2 n$ (valid when these terms exist; for small $n$ check directly).

AP condition $2 c_{2} = c_{1} + c_{3}$ :

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

Simplifying: $n^{3} - 9 n^{2} + 26 n - 24 = 0$ , which factors as $(n - 2) (n - 3) (n - 4) = 0$ .

Each of $n = 2, 3, 4$ satisfies the AP condition (verifiable directly). Sum $= 2 + 3 + 4 = 9$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →