Question

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

Answer 1

Using binomial theorem for the expansion of ${(\frac{2}{x}-\frac{x}{2})}^5$, we have:

${(\frac{2}{x}-\frac{x}{2})}^5=5_{C_0}{\left(\frac{2}{x}\right)}^5+5_{C_1}{\left(\frac{2}{x}\right)}^4\left(\frac{-x}{2}\right)+5_{C_2}{\left(\frac{2}{x}\right)}^3{\left(\frac{-x}{2}\right)}^2+5_{C_3}{\left(\frac{2}{x}\right)}^2{\left(\frac{-x}{2}\right)}^3+5_{C_4}\left(\frac{2}{x}\right){\left(\frac{-x}{2}\right)}^4+5_{C_5}{\left(\frac{-x}{2}\right)}^5$

$=\frac{32}{x^5}+5.\frac{16}{x^4}.\frac{-x}{2}+10.\frac{8}{x^3}.\frac{x^2}{4}+10.\frac{4}{x^2}.\frac{-x^3}{8}+5.\frac{2}{x}.\frac{x^4}{16}+\frac{-x^5}{32}$

$=\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5}{8}{x}^3-\frac{x^5}{32}$