Answer to Question #272188 in Calculus for Mimi

Question

Find the fourth term of the binomial expansion (x-2y²)⁵

Answer 1

$=\frac{5}{0!(5-0)!}x^5(-2y^2)^0+\frac{5}{1!(5-1)!}x^4(-2y^2)^1+\frac{5}{2!(5-2)!}x^3(-2y^2)^2+\frac{5}{3!(5-3)!}x^2(-2y^2)^3\\$

$=x^5-10x^4y^2+40x^3y^4-80x^2y^6$