JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The general term in the expansion of ( 1 x^3 - x^4 )^n is given by: T_ r+1 = ^ n C_ r (x^ -3 )^ n-r (-x^4 )^r = (-1)^r ^ n C_ r x^ 7r - 3n For the coefficient of x^7, we set the exponent to 7: 7r_1 - 3n = 7 r_1 = 3n + 7 7 For the coefficient of x^ 14 , we set the exponent to 14: 7r_2 - 3n = 14 r_2 = 3n + 14 7 Notice that r_2 = r_1 + 1. The sum of the coefficients of x^7 and x^ 14 is zero: (-1)^ r_1 ^ n C_ r_1 + (-1)^ r_2 ^ n C_ r_2 = 0 (-1)^ r_1 ^ n C_ r_1 + (-1)^ r_1 + 1 ^ n C_ r_1 + 1 = 0 (-1)^ r_1 ( ^ n C_ r_1 - ^ n C_ r_1 + 1 ) = 0 ^ n C_ r_1 = ^ n C_ r_1 + 1 Since r_1 r_1 + 1, we must use the property ^ n C_ x = ^ n C_ y x + y = n: r_1 + (r_1 + 1) = n 2r_1 + 1 = n Substituting r_1 = 3n + 7 7 : 2 ( 3n + 7 7 ) + 1 = n 6n + 14 7 + 1 = n 6n + 14 + 7 = 7n n = 21 Answer: 21