If the coefficient of x^7 in the binomial expansion of (ax - 1 bx^2 )^n and the coefficient of x⁻⁵ in the binomial expansion of (ax + 1 bx^2 )^n are equal, and a^4 b^4 = 22 , then the positive integer n is equal to:

Step-by-step solution

Let the general term in the expansion of (ax - 1 bx^2 )^n be T_ r₁+1 . T_ r₁+1 = ^ n C_ r₁ (ax)^ n-r₁ (- 1 bx^2 )^ r₁ = (-1)^ r₁ ^ n C_ r₁ a^ n-r₁ b^ -r₁ x^ n-3r₁ For the coefficient of x^7 , we set n - 3r₁ = 7 n = 3r₁ + 7 . Similarly, let the general term in the expansion of (ax + 1 bx^2 )^n be T_ r₂+1 . T_ r₂+1 = ^ n C_ r₂ (ax)^ n-r₂ ( 1 bx^2 )^ r₂ = ^ n C_ r₂ a^ n-r₂ b^ -r₂ x^ n-3r₂ For the coefficient of x⁻⁵ , we set n - 3r₂ = -5 n = 3r₂ - 5 . Equating the two expressions for n , we get 3r₁ + 7 = 3r₂ - 5 r₂ - r