# How do you use the binomial theorem to expand and simplify the expression (1/x+y)^5?

Jan 29, 2018

$\textcolor{red}{{\left(\frac{1}{x} + y\right)}^{5} = \frac{1}{x} ^ 5 + \frac{5 y}{x} ^ 4 + \frac{10 {y}^{2}}{x} ^ 3 + \frac{10 {y}^{3}}{x} ^ 2 + \frac{5 {y}^{4}}{x} + {y}^{5}}$

#### Explanation:

As per Pascal's 5th row, constant coefficients of the 6 terms are

## 1 5 10 10 5 1

${\left(\frac{1}{x} + y\right)}^{5} = \left(1 \cdot {\left(\frac{1}{x}\right)}^{5} {y}^{0} + 5 \cdot {\left(\frac{1}{x}\right)}^{4} y + 10 \cdot {\left(\frac{1}{x}\right)}^{3} {y}^{2} + 10 \cdot \left(\frac{1}{x} 2\right) {y}^{3} + 5 \cdot \left(\frac{1}{x}\right) \cdot {y}^{4} + 1 \cdot {\left(\frac{1}{x}\right)}^{0} \cdot {y}^{5}\right)$

(1/x+ y)^5 = x^-5 + 5 x^-4 y + 10 x^-3 y^2 + 10 x_-2 y^3 + 5 x^-1 y^4 + y^)

$\textcolor{red}{{\left(\frac{1}{x} + y\right)}^{5} = \frac{1}{x} ^ 5 + \frac{5 y}{x} ^ 4 + \frac{10 {y}^{2}}{x} ^ 3 + \frac{10 {y}^{3}}{x} ^ 2 + \frac{5 {y}^{4}}{x} + {y}^{5}}$