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 :

Mathematics · Binomial Theorem

JEE Main 2026 — 23 January, Morning Shift — Question 14

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

Option A:
3
Option B:
7
Option C:
12
Option D:
9
Correct

Answer: D

Step-by-step solution

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

Coefficient $\mathrm{x} \Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{1}$ ,

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

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

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

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

Now according to question $\mathrm{n}+2 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=2\left[2+\frac{\mathrm{n}(\mathrm{n}-1)}{2}\right]$

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

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

$\Rightarrow \mathrm{n}=2,3,4 \quad \Rightarrow \mathrm{n}=3,4$

Now checking for $\mathrm{n}=2$

Coeff. of $x=2$

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

$\Rightarrow \mathrm{n}=2$ is also the correct choice

Required sum of values of ' n ' $=2+3+4=9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients