Question

Expand using binomial theorem
${[1+\frac{x}{2}-\frac{2}{x}]}^4,x\ne 0$.

Answer 1

We have ${[1+\frac{x}{2}-\frac{2}{x}]}^4$

$={[1+(\frac{x}{2}-\frac{2}{x})]}^4$

$=4_{C_0}+4_{C_1}(\frac{x}{2}-\frac{2}{x})+4_{C_2}{(\frac{x}{2}-\frac{2}{x})}^2+4_{C_3}{(\frac{x}{2}-\frac{2}{x})}^3+4_{C_4}{(\frac{x}{2}-\frac{2}{x})}^4$

$=1+4(\frac{x}{2}-\frac{2}{x})+6(\frac{x^2}{4}+\frac{4}{x^2}-2)+4(\frac{x^3}{8}-\frac{8}{x^3}-\frac{3x}{2}+\frac{6}{x})+[4_{C_0}{\left(\frac{x}{2}\right)}^2{\left(\frac{2}{x}\right)}^2-4_{C_3}\left(\frac{x}{2}\right){\left(\frac{2}{x}\right)}^3+4_{C_4}{\left(\frac{2}{x}\right)}^4]$

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

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