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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Binomial Theorem

The sum of all possible values of n ∈ N , so that the coefficients...

The sum of all possible values of n∈N\mathrm{n} \in \mathbf{N}, so that the coefficients of x,x2andx3intheexpansionof(1+x2)2(1+x)nx, x^{2} and x^{3} in the expansion of \left(1+x^{2}\right)^{2}(1+x)^{\mathrm{n}}, are in arithmetic progression is :

Options

  1. A12
  2. B7
  3. C3
  4. D9

Correct answer

D. 9

Step-by-step solution

(1+x2)2=1+2x2+x4(1+x^2)^2 = 1 + 2x^2 + x^4. Multiply with (1+x)n(1+x)^n:
Coefficient of xx: nn.
Coefficient of x2x^2: (n2)+2=n(n−1)2+2\binom{n}{2} + 2 = \frac{n(n-1)}{2} + 2.
Coefficient of x3x^3: (n3)+2n=n(n−1)(n−2)6+2n\binom{n}{3} + 2n = \frac{n(n-1)(n-2)}{6} + 2n.
For AP: 2(n(n−1)2+2)=n+n(n−1)(n−2)6+2n2\left(\frac{n(n-1)}{2}+2\right) = n + \frac{n(n-1)(n-2)}{6} + 2n.
Simplifying: n3−9n2+26n−24=0n^3 - 9n^2 + 26n - 24 = 0.
(n−2)(n−3)(n−4)=0⇒n=2,3,4(n-2)(n-3)(n-4) = 0 \Rightarrow n = 2, 3, 4.
Sum =2+3+4=9= 2 + 3 + 4 = 9.
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