Question

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

Answer 1

$(1+\frac{x}{2}-\frac{2}{x})$

$(a+b{)}^n{=}^n{C}_0{a}^n{b}^0{+}^n{C}_1{a}^{n-1}{b}^1{+}^m{C}_2{a}^{n-2}{b}^2+........{.}^n{C}_n{a}^n{b}^n$

Hence

$(a+b{)}^4{=}^4{C}_0{a}^4(b^0){+}^4{C}_1{a}^3{b}^1{+}^4{C}_2{a}^2{b}^2{+}^4{C}_{3}a.b^3{+}^4{C}_4{a}^n{b}^4$

by expanding $\Rightarrow a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$----- $\left(1\right)$

${(1+\frac{x}{2}-\frac{2}{x})}^4\Rightarrow a=1+\frac{x}{2}b=-\frac{2}{x}$

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

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

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

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

Putting $\left(3\right)8\left(4\right)$ in $\left(2\right)$

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

by solving

$=\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^9}$

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