Question

Expand the expression
${(\frac{2}{x}-\frac{x}{2})}^5$

Answer 1

$(x+a{)}^n={\sum }_{r=0}^n{}^n{\text{C}}_r{x}^r{a}^{n-r}$

${(\frac{2}{x}-\frac{x}{2})}^5=\sum _{r=0}^5{}^5{C}_r{\left(\frac{2}{x}\right)}^r{(-\frac{x}{2})}^{5-r}$
${=}^5{C}_0{\left(\frac{2}{x}\right)}^0{(-\frac{x}{2})}^5{+}^5{C}_1\left(\frac{2}{x}\right){(-\frac{x}{2})}^4{+}^5{C}_2{\left(\frac{2}{x}\right)}^2{(-\frac{x}{2})}^3{+}^5{C}_3{\left(\frac{2}{x}\right)}^3{(-\frac{x}{2})}^2{+}^5{C}_4{\left(\frac{2}{x}\right)}^4{(-\frac{x}{2})}^1{+}^5{C}_5{\left(\frac{2}{x}\right)}^5{(-\frac{x}{2})}^0$
$={-}^5{C}_0\left(\frac{x^5}{32}\right){+}^5{C}_1\left(\frac{x^3}{8}\right){-}^5{C}_2\left(\frac{x}{2}\right){+}^5{C}_3\left(\frac{2}{x}\right){-}^5{C}_4\left(\frac{8}{x^4}\right){+}^5{C}_5\left(\frac{32}{x^5}\right)$