Precalculus, Chapter 9, 9.5, Section 9.5, Problem 32

You need to use the binomial formula, such that:
(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k
You need to replace 2x for x, 5y for y and 5 for n, such that:
(2x - 5y)^5 = 5C0 (2x)^5+5C1 (2x)^4*(-5y)^1+5C2 (2x)^3*(-5y)^2+5C3 (2x)^2*(-5y)^3 + 5C4 2x*(-5y)^4 + 5C5 (-5y)^5
By definition, nC0 = nCn = 1, hence 5C0 = 5C5 = 1.
By definition nC1 = nC(n-1) = n, hence 5C1 = 5C4 = 5.
By definition nC2 = (n(n-1))/2 , hence 5C2 = 5C3 = 10.
(2x - 5y)^5 = 32x^5-400x^4*y+2000x^3*y^2-5000x^2*y^3 + 6250x*y^4 -3125y^5
Hence, expanding the complex number using binomial theorem yields the simplified result (2x - 5y)^5 = 32x^5-400x^4*y+2000x^3*y^2-5000x^2*y^3 + 6250x*y^4 -3125y^5.