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BINOMIAL THEOREM — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBINOMIAL THEOREM3 Marks
Question
Expand the given expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$
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Answer

Using binomial theorem for the expansion of ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ we have
${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$${ = ^5}{C_0}{\left( {\frac{x}{3}} \right)^5}{ + ^5}{C_1}{\left( {\frac{x}{3}} \right)^4}\left( {\frac{1}{x}} \right)$${ + ^5}{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2}{ + ^5}{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3}$
${ + ^5}{C_4}\left( {\frac{x}{3}} \right){\left( {\frac{1}{x}} \right)^4}{ + ^5}{C_5}{\left( {\frac{1}{x}} \right)^5}$
$ = \frac{{{x^5}}}{{243}} + 5 \cdot \frac{{{x^4}}}{{81}} \cdot \frac{1}{x} + 10 \cdot \frac{{{x^3}}}{{27}} \cdot \frac{1}{{{x^2}}}$$ + 10 \cdot \frac{{{x^2}}}{9} \cdot \frac{1}{{{x^3}}} + 5 \cdot \frac{x}{3} \cdot \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}$
$ = \frac{{{x^5}}}{{243}} + \frac{5}{{81}}{x^3} + \frac{{10}}{{27}}x + \frac{{10}}{{9x}} + \frac{5}{{3{x^3}}} + \frac{1}{{{x^5}}}$

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