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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Expand the expression: $\large(\frac{2}{x} - \frac{x}{2})$$^5$

$\begin{array}{1 1} \large\frac{32}{x^5} - \large\frac{40}{x^3} + \large\frac{20}{x} -5x+\large\frac{5x}{8}-\large\frac{x^5}{32} \\ \large\frac{32}{x^5} + \large\frac{40}{x^3} - \large\frac{20}{x} -5x+\large\frac{5x}{8}-\large\frac{x^5}{32} \\ \large\frac{32}{x^5} - \large\frac{40}{x^3} + \large\frac{20}{x} +5x-\large\frac{5x}{8}+\large\frac{x^5}{32} \\ 32 - \large\frac{40}{x^3} - \large\frac{20}{x} -5x+\large\frac{5x}{8}-\large\frac{x^5}{32} \end{array} $

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  • $(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$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
Given $\large(\frac{2}{x} - \frac{x}{2})$$^5 \rightarrow a = \large\frac{2}{x}$ and $b = \large\frac{-x}{2}$
$\Rightarrow \large(\frac{2}{x} - \frac{x}{2})$$^5 = \large \sum \limits_{k=0}^{5}\; $$^5 \large C$$_k \; a^{5-k}b^k $
$\qquad = \; ^5 \large C$$_0 \; (2/x)^{5-0}(-x/2)^0 +\; ^5 \large C$$_1 \; (2/x)^{5-1}(-x/2)^{1} + ^5 \large C$$_2 \; (2/x)^{5-2}(-x/2)^2 + ^5 \large C$$_3 \; (2/x)^{5-3}(-x/2)^3 + ^5 \large C$$_4 \; (2/x)^{5-4}(-x/2)^4 + ^5 \large C$$_5 \; (2/x)^{5-5}(-x/2)^5$
$\qquad = \large\frac{32}{x^5} $$-5\large(\frac{16}{x4}\;\frac{x}{2})$$+10\large(\frac{8}{x^3}\frac{x^2}{4})$$-10\large(\frac{4}{x^2}\frac{x^3}{8})$$+5\large(\frac{2}{x}\frac{x^4}{16})$$ - \large\frac{x^5}{32}$
$\qquad = \large\frac{32}{x^5}$$ - \large\frac{40}{x^3}$$ + \large\frac{20}{x}$$ -5x+\large\frac{5x}{8}$$-\large\frac{x^5}{32}$
answered Apr 9, 2014 by balaji.thirumalai
edited Apr 9, 2014 by balaji.thirumalai
 

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