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The term independent of $x$ in the expansion $(2x-\large\frac{1}{3x})^6$ is ?

$\begin{array}{1 1} \frac{160}{9} \\ \frac{80}{9} \\ \frac{160}{27} \\ \frac{80}{3} \end{array}$

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1 Answer

General term $T_{r+1}$ in the expansion of $(2x-\large\frac{1}{3x})^6$ is $(-1)^r.^6C_r.(2x)^{6-r}.(\large\frac{1}{3x})^r$
$=(-1)^r.^6C_r.2^{6-r}.3^{-r}x^{6-2r}$
For the term independent of $x$, $6-2r=0$
$\Rightarrow\:r=3$
The term is $-^6C_3.2^3.3^{-3}$
$=-20\times 8\times - \large\frac{1}{27}$
$=\large\frac{160}{27}$

 

answered Sep 17, 2013 by rvidyagovindarajan_1
edited Dec 23, 2013 by meenakshi.p
 

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