Find the fourth term of the binomial expansion (x-2y2)5
=50!(5−0)!x5(−2y2)0+51!(5−1)!x4(−2y2)1+52!(5−2)!x3(−2y2)2+53!(5−3)!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\\=0!(5−0)!5x5(−2y2)0+1!(5−1)!5x4(−2y2)1+2!(5−2)!5x3(−2y2)2+3!(5−3)!5x2(−2y2)3
=x5−10x4y2+40x3y4−80x2y6=x^5-10x^4y^2+40x^3y^4-80x^2y^6=x5−10x4y2+40x3y4−80x2y6
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