If ^ 2n C₃ : ^ n C₂ = 44 : 3 , then the coefficient of x⁴ in the expansion of (1 - 2x)^ n is

Step-by-step solution

Given that ^ 2n C₃ : ^ n C₂ = 44 : 3 . Expanding the combinations, we get: 2n(2n-1)(2n-2) 3! n(n-1) 2! = 44 3 2n(2n-1) 2(n-1) 6 2 n(n-1) = 44 3 4n(2n-1)(n-1) 3 1 n(n-1) = 44 3 Cancelling n(n-1) from the numerator and denominator (since n 2 ): 4(2n-1) 3 = 44 3 4(2n-1) = 44 2n-1 = 11 2n = 12 n = 6 We need to find the coefficient of x⁴ in the expansion of (1 - 2x)⁶ . The general term in the expansion of (1 - 2x)⁶ is T_ r+1 = ⁶C_ r (-2x)^ r . For the coefficient of x⁴ , we put r = 4 : Coefficient = ⁶C₄ (-2)⁴ Coefficien