Question

Expand using Binomial Theorem ${(1+\frac{x}{2}-\frac{2}{x})}^4,x\ne 0$.

Answer 1

$\begin{array}{c}{\left(1+\frac{x}{2}-\frac{2}{x}\right)}^4\\ ={\left[\left(\frac{x}{2}-\frac{2}{x}\right)+1\right]}^4\\ {=}^4{C}_0{\left(\frac{x}{2}-\frac{2}{x}\right)}^4{\left(1\right)}^0{+}^4{C}_1{\left(\frac{x}{2}-\frac{2}{x}\right)}^3{\left(1\right)}^1{+}^4{C}_2{\left(\frac{x}{2}-\frac{2}{x}\right)}^2{\left(1\right)}^2{+}^4{C}_3{\left(\frac{x}{2}-\frac{2}{x}\right)}^1{\left(1\right)}^3{+}^4{C}_4{\left(\frac{x}{2}-\frac{2}{x}\right)}^0{\left(1\right)}^4\\ ={\left(\frac{x}{2}-\frac{2}{x}\right)}^4+4{\left(\frac{x}{2}-\frac{2}{x}\right)}^3+6{\left(\frac{x}{2}-\frac{2}{x}\right)}^2+4\left(\frac{x}{2}-\frac{2}{x}\right)+1\\ =\left({}^4{C}_0{\left(\frac{x}{2}\right)}^4{\left(\frac{2}{x}\right)}^0{-}^4{C}_1{\left(\frac{x}{2}\right)}^3{\left(\frac{2}{x}\right)}^1{+}^4{C}_2{\left(\frac{x}{2}\right)}^2{\left(\frac{2}{x}\right)}^2{-}^4{C}_3{\left(\frac{x}{2}\right)}^1{\left(\frac{2}{x}\right)}^3{+}^4{C}_4{\left(\frac{x}{2}\right)}^0{\left(\frac{2}{x}\right)}^4\right)\\ +4\left({}^3{C}_0{\left(\frac{x}{2}\right)}^3{\left(\frac{2}{x}\right)}^0{-}^3{C}_1{\left(\frac{x}{2}\right)}^2{\left(\frac{2}{x}\right)}^1{+}^3{C}_2{\left(\frac{x}{2}\right)}^1{\left(\frac{2}{x}\right)}^2{-}^3{C}_3{\left(\frac{x}{2}\right)}^0{\left(\frac{2}{x}\right)}^3\right)+6\left(\frac{x^2}{4}+\frac{4}{x^2}-2\right)+2x-\frac{8}{x}+1\\ =\frac{x^4}{16}-x^2+6-\frac{16}{x^2}+\frac{16}{x^4}+\frac{x^3}{2}-6x+\frac{24}{x}-\frac{32}{x^3}+\frac{3}{2}{x}^2+\frac{24}{x^2}-12+2x-\frac{8}{x}+1\\ =\frac{x^4}{16}+\frac{x^3}{2}+\frac{1}{2}{x}^2-4x+\frac{16}{x^4}-\frac{32}{x^3}+\frac{8}{x^2}+\frac{16}{x}-5\end{array}$