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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Expand each of the expressions:$\big(\large\frac{x}{3}+\frac{1}{x}\big)^5$

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$\big(\large\frac{x}{3}+\frac{1}{x}\big)^5$
$\Rightarrow 5C_0(\large\frac{x}{3})^5(\frac{1}{x})^0$$+5C_1(\large\frac{x}{3})^4(\frac{1}{x})$$+5C_2(\large\frac{x}{3})^3(\frac{1}{x})^2$$+5C_3(\large\frac{x}{3})^2(\frac{1}{x})^3$$+5C_4(\large\frac{x}{3})(\frac{1}{x})^4$$+5C_5(\large\frac{x}{3})(\frac{1}{x})^5$
$\Rightarrow \large\frac{x^5}{243}+\frac{5}{1}.\frac{x^4}{81}.\frac{1}{x}+\frac{5.4}{1.2}.\frac{x^3}{27}.\frac{1}{x^2}+\frac{5.4}{1.2}.\frac{x^2}{9}.\frac{1}{x^3}+\frac{5}{1}.\frac{x}{3}.\frac{1}{x^4}+\frac{1}{x^5}$
$\Rightarrow \large\frac{x^5}{243}+\frac{5}{81}$$.x^3+\large\frac{10}{27}$$.x+\large\frac{10}{9}.\frac{1}{x}+\frac{5}{3}.\frac{1}{x^3}+\frac{1}{x^5}$
answered Jun 20, 2014 by sreemathi.v
 
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