દ્વિપદી પ્રમેયનો ઉપયોગ કરી $\left(1+\frac{ x }{2}-\frac{2}{ x }\right)^{4}, x \neq 0$ નું વિસ્તરણ કરો. 

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

$ = {\,^n}{C_0}{\left( {1 + \frac{x}{2}} \right)^4} - {\,^n}{C_1}{\left( {1 + \frac{x}{2}} \right)^3}\left( {\frac{2}{x}} \right) + {\,^n}{C_2}{\left( {1 + \frac{x}{2}} \right)^2}{\left( {\frac{2}{x}} \right)^2}$

            $ - {\,^n}{C_3}\left( {1 + \frac{x}{2}} \right){\left( {\frac{2}{x}} \right)^3} + {\,^n}{C_4}{\left( {\frac{2}{x}} \right)^4}$

$=\left(1+\frac{x}{2}\right)^{4}-4\left(1+\frac{x}{2}\right)^{3}\left(\frac{2}{x}\right)+6\left(1+x+\frac{x^{2}}{4}\right)\left(\frac{4}{x^{2}}\right)-4\left(1+\frac{x}{2}\right)\left(\frac{8}{x^{3}}\right)+\frac{16}{x^{4}}$

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

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

Again by using Binomial Theorem, we obtain

${\left( {1 + \frac{x}{2}} \right)^4} = {\,^4}{C_0}{(1)^4} + {\,^4}{C_1}{(1)^3}\left( {\frac{x}{2}} \right) + {\,^4}{C_2}{(1)^2}{\left( {\frac{x}{2}} \right)^2}$

                 $ + {\,^4}{C_3}{(1)^3}{\left( {\frac{x}{2}} \right)^3} + {\,^4}{C_4}{\left( {\frac{x}{2}} \right)^4}$

$=1+4 \times \frac{x}{2}+6 \times \frac{x^{4}}{4}+4 \times \frac{x^{3}}{8}+\frac{x^{4}}{16}$

$=1+2 x+\frac{3 x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}$         .........$(2)$

${\left( {1 + \frac{x}{2}} \right)^3} = {\,^3}{C_0}{(1)^3} + {\,^3}{C_1}{(1)^2}\left( {\frac{x}{2}} \right) + {\,^3}{C_2}(1)\left( {\frac{x}{2}} \right) + {\,^3}{C_3}{\left( {\frac{x}{2}} \right)^3}$

$=1+\frac{3 x}{2}+\frac{3 x^{2}}{4}+\frac{x^{3}}{8}$          ..........$(3)$

From $(1), (2)$ and $(3),$ we obtain

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

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

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

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