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

$\begin{array}{1 1} \large\frac{x^5}{243} + \large\frac{5x^3}{81} + \large\frac{10x}{27} +\large\frac{10}{9x}+\large\frac{5}{3x^3}+\large\frac{1}{x^5} \\ \large\frac{243}{x^5} + \large\frac{81}{5x^3} + \large\frac{27}{10x} +\large\frac{9x}{10}+\large\frac{3x^3}{5}+\large\frac{x^5}{1} \\ \large\frac{81}{x^5} + \large\frac{27}{5x^3} + \large\frac{9}{10x} +\large\frac{x}{10}+\large\frac{9x^3}{5}+\large\frac{x^5}{3} \\ \large\frac{x^5}{81} + \large\frac{5x^3}{27} + \large\frac{10x}{9} +\large\frac{10}{3x}+\large\frac{5}{x^3}+\large\frac{1}{x^5} \end{array} $

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1 Answer

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Toolbox:
  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
Given $\large(\frac{x}{3} + \frac{1}{x})$$^5 \rightarrow a = \large\frac{x}{3}$ and $b = \large\frac{1}{x}$
$\Rightarrow \large(\frac{x}{3} $$+ \frac{1}{x})$$^5 = \large \sum \limits_{k=0}^{5}\; $$^5 \large C$$_k \; (x/3)^{5-k}(1/x)^k $
$\qquad = \; ^5 \large C$$_0 \; (x/3)^{5-0}(1/x)^0 +\; ^5 \large C$$_1 \; (x/3)^{5-1}(1/x)^{1} + ^5 \large C$$_2 \; (x/3)^{5-2}(1/x)^2 + ^5 \large C$$_3 \; (x/3)^{5-3}(1/x)^3 + ^5 \large C$$_4 \; (x/3)^{5-4}(1/x)^4 + ^5 \large C$$_5 \; (x/3)^{5-5}(1/x)^5$
$\qquad = \large\frac{x^5}{243} $$+5\large(\frac{x^4}{81}\;\frac{1}{x})$$+10\large(\frac{x^3}{27}\frac{1}{x})$$+10\large(\frac{x^2}{9}\frac{1}{x^3})$$+5\large(\frac{x}{3}\frac{1}{x^4})$$ +\large\frac{1}{x^5}$
$\qquad = \large\frac{x^5}{243}$$ + \large\frac{5x^3}{81}$$ + \large\frac{10x}{27}$$ +\large\frac{10}{9x}$$+\large\frac{5}{3x^3}$$+\large\frac{1}{x^5}$
answered Apr 9, 2014 by balaji.thirumalai
edited Apr 9, 2014 by balaji.thirumalai
 

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