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Binomial Theorem — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSBinomial Theorem3 Marks
Question
Expand the given expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$
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Answer

Using binomial theorem for the expansion of ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$ we have
${\left( {\frac{2}{x} - \frac{x}{2}} \right)^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 \cdot \frac{{16}}{{{x^4}}} \cdot \frac{{ - x}}{2} + 10 \cdot \frac{8}{{{x^3}}} \cdot \frac{{{x^2}}}{4}$$ + 10 \cdot \frac{4}{{{x^2}}} \cdot \frac{{ - {x^3}}}{8} + 5 \cdot \frac{2}{x} \cdot \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}}$

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