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BINOMIAL THEOREM — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBINOMIAL THEOREM3 Marks
Question
Expand using binomial theorem ${\left[ {1 + \frac{x}{2} - \frac{2}{x}} \right]^4},x \ne 0$
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Answer

We have ${\left[ {1 + \frac{x}{2} - \frac{2}{x}} \right]^4} = {\left[ {1 + \left( {\frac{x}{2} - \frac{2}{x}} \right)} \right]^4}$
${ = ^4}{C_0}{ + ^4}{C_1}\left( {\frac{x}{2} - \frac{2}{x}} \right){ + ^4}{C_2}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^2}$${ + ^4}{C_3}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^3}{ + ^4}{C_4}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^4}$
$ = 1 + 4\left( {\frac{x}{2} - \frac{2}{x}} \right) + 6\left( {\frac{{{x^2}}}{4} + \frac{4}{{{x^2}}} - 2} \right)$$ + 4\left( {\frac{{{x^3}}}{8} - \frac{8}{{{x^3}}} - \frac{{3x}}{2} + \frac{6}{x}} \right)$
$ + \left[ {^4{C_0}{{\left( {\frac{x}{2}} \right)}^4}{ - ^4}{C_1}{{\left( {\frac{x}{2}} \right)}^3}\left( {\frac{2}{x}} \right){ + ^4}{C_2}} \right.$${\left( {\frac{x}{2}} \right)^2}$$\left. {{{\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}} \right]$
$ = 1 + \left( {2x - \frac{8}{x}} \right) + \left( {\frac{3}{2}{x^2} + \frac{{24}}{{{x^2}}} - 12} \right)$$ + \left( {\frac{{{x^3}}}{2} - \frac{{32}}{{{x^3}}} - 6x + \frac{{24}}{x}} \right)$
$ + \left( {\frac{{{x^4}}}{{16}} - {x^2} + 6 - \frac{{16}}{{{x^2}}} + \frac{{16}}{{{x^4}}}} \right)$
$ = - 5 - 4x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^4}}}{{16}} + \frac{{16}}{x} + \frac{8}{{{x^2}}}$$ - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}}$

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