1
The sum of all possible values of n N, so that the coefficients of x, x2 and x3 in the expansion of (1 + x2)2(1 + x)n, are in arithmetic progression is :
- 12
- 9
- 3
- 7
Answer
(2) 9
Solution
(1 + x2)2(1 + x)n = (1 + 2x2 + x4)(1 + x)n
Coefficient of x: nC1 = n
Coefficient of x2: nC2 + 2
Coefficient of x3: nC3 + 2 nC1 = nC3 + 2n
In A.P. 2(nC2 + 2) = n + nC3 + 2n
n(n - 1) + 4 = 3n + n(n - 1)(n - 2)6
6n2 - 6n + 24 = 18n + n3 - 3n2 + 2n
n3 - 9n2 + 26n - 24 = 0
(n - 2)(n - 3)(n - 4) = 0 n = 2, 3, 4
(Check n = 2: coefficients 2, 3, 4 are in A.P.)
Sum = 2 + 3 + 4 = 9
Answer: (2)
JEE Main 2026 (Online) 23rd January Morning Shift