Question

Expand using Binomial Theorem $( 1+ \frac{X}{2} - \frac{2}{X})^4, X≠0$

Answer 1

Using Binomial Theorem, the given expression $( 1 +\frac{ x}{2} - \frac{2}{x})^4$ can be expanded as
$[ ( 1 +\frac{ x}{2} )- \frac{2}{x} ]^4$

$=\space^ 4C_0(1 + \frac{x}{2})^4 - \space^4C_1(1 +\frac{ x}{2})^3(\frac{2}{x}) + \space^4C_2(1 +\frac{ x}{2})^2(\frac{2}{x})^2 - \space^4C_3(1 +\frac{ x}{2}) (\frac{2}{x})^3 + \space^4C_4(\frac{2}{x})^4$

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

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

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

Again by using Binomial Theorem, we obtain

$(1 +\frac{ x}{2})^4 = \space^3C_0 (1)^4+\space^ 4C_1(1)^3(\frac{x}{2})+ \space ^4C_2(1)^2(\frac{x}{2})^2 \+ \space^4C_3(1)(\frac{x}{2})^3 \+ \space^4C_4(\frac{x}{2})^4$ $=1+4 \times\frac{x}{2}+6\times \frac{x^2}{4} + 4 \times \frac{x^3}{8} +\frac{x^4}{16} ...(2)$

$(1 + \frac{x}{2})^3= \space^3C_0(1 +\frac{ x}{2})^3 - \space^3C_1(1 )^2 (\frac{x}{2})+\space^ 3C_2(1) (\frac{x}{2})^2 - \space^3C_3( \frac{x}{2})^3$ $=1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8} ....(3)$

from (1), (2) and (3), we obtain
$[(1 + \frac{x}{2}) - \frac{2}{x}]^4$

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

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

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