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JEE Main20265 April 2026Morning ShiftMathematicsBinomial TheoremActual

If the sum of the coefficients of x^7 and x¹⁴ in the expansion of ( 1 x^3 - x^4 )^n , x 0 , is zero, then the value of n is __________.

Correct answer

0

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 : 7r₁ - 3n = 7 r₁ = 3n + 7 7 For the coefficient of x¹⁴ , we set the exponent to 14 : 7r₂ - 3n = 14 r₂ = 3n + 14 7 Notice that r₂ = r₁ + 1 . The sum of the coefficients of x^7 and x¹⁴ is zero: (-1)^ r₁ ^ n C_ r₁ + (-1)^ r₂ ^ n C_ r₂ = 0 (-1)^ r₁ ^ n C_ r₁ + (-1)^ r₁ + 1 ^ n C_ r₁ + 1 = 0 (-1)^ r₁ ( ^ n C_ r₁ - ^ n C_ r₁ + 1 ) =

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