Question

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

Answer 1

${(1+\frac{x}{2}-\frac{2}{x})}^4={((1+\frac{x}{2})-\frac{2}{x})}^4={(1+\frac{x}{2})}^4+4{(1+\frac{x}{2})}^3\left(\frac{-2}{x}\right)+6{(1+\frac{x}{2})}^2{\left(\frac{-2}{x}\right)}^2+4(1+\frac{x}{2}){\left(\frac{-2}{x}\right)}^3+{\left(\frac{-2}{x}\right)}^4$

$={(1+\frac{x}{2})}^4-\frac{8}{x}{(1+\frac{x}{2})}^3+\frac{24}{x^2}{(1+\frac{x}{2})}^2-\frac{32}{x^3}(1+\frac{x}{2})+\frac{16}{x^4}$

$\{1^4+{4.1}^3.\frac{x}{2}+{6.1}^2.{\left(\frac{x}{2}\right)}^3+\mathrm{4.1.}{\left(\frac{x}{2}\right)}^3{\left(\frac{x}{2}\right)}^4\}-\frac{8}{x}\{1^3+{3.1}^2.\frac{x}{2}+\mathrm{3.1.}{\left(\frac{x}{2}\right)}^2+{\left(\frac{x}{2}\right)}^3\}+\frac{24}{x^2}\{1+\frac{x^2}{4}+x\}-\frac{32}{x^3}(1+\frac{x}{2})+\frac{16}{x^4}$

$=1+2x+\frac{3}{2}{x}^2+\frac{x^3}{2}+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2+\frac{24}{x^2}+6+\frac{24}{x}-\frac{32}{x^3}-\frac{16}{x^2}+\frac{16}{x^4}$

$=\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{3}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$

$\therefore {(1+\frac{x}{2}-\frac{2}{x})}^4=\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$