How do you use the binomial theorem to expand and simplify the expression #(1/x+y)^5#? Precalculus The Binomial Theorem The Binomial Theorem 1 Answer Cem Sentin Dec 19, 2017 #(1/x+y)^5=1/(x^5)+(5y)/x^4+(10y^2)/x^3+(10y^3)/x^2+(5y^4)/x+y^5# Explanation: #(1/x+y)^5# =#(1/x)^5y^0*C(5,0)+(1/x)^4y^1*C(5,1)+(1/x)^3y^2*C(5,2)+(1/x)^2y^3*C(5,3)+(1/x)^1y^4*C(5,4)+(1/x)^0y^5*C(5,5)# =#1/(x^5)+(5y)/x^4+(10y^2)/x^3+(10y^3)/x^2+(5y^4)/x+y^5# Answer link Related questions What is the binomial theorem? How do I use the binomial theorem to expand #(d-4b)^3#? How do I use the the binomial theorem to expand #(t + w)^4#? How do I use the the binomial theorem to expand #(v - u)^6#? How do I use the binomial theorem to find the constant term? How do you find the coefficient of x^5 in the expansion of (2x+3)(x+1)^8? How do you find the coefficient of x^6 in the expansion of #(2x+3)^10#? How do you use the binomial series to expand #f(x)=1/(sqrt(1+x^2))#? How do you use the binomial series to expand #1 / (1+x)^4#? How do you use the binomial series to expand #f(x)=(1+x)^(1/3 )#? See all questions in The Binomial Theorem Impact of this question 1367 views around the world You can reuse this answer Creative Commons License