Precalculus, Chapter 9, 9.5, Section 9.5, Problem 35

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 1/x for x, y for y and 5 for n, such that:
(1/x + y)^5 = 5C0 (1/x)^5+5C1 (1/x)^4*(y)^1+5C2 (1/x)^3*(y)^2+5C3 1/(x^2)*(y)^3 + 5C4 1/x*(y)^4 + 5C5 (y)^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.
(1/x + y)^5 = 1/(x^5)+(5y)/(x^4)+ (10y^2)1/(x^3)+(10y^3)/(x^2) + (5y^4)/x + (y)^5
Hence, expanding the complex number using binomial theorem yields the simplified result (1/x + y)^5 = 1/(x^5)+(5y)/(x^4)+ (10y^2)1/(x^3)+(10y^3)/(x^2) + (5y^4)/x + (y)^5.