Question #272188

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

Expert's answer

=50!(50)!x5(2y2)0+51!(51)!x4(2y2)1+52!(52)!x3(2y2)2+53!(53)!x2(2y2)3=\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\\



=x510x4y2+40x3y480x2y6=x^5-10x^4y^2+40x^3y^4-80x^2y^6


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