The term independent of ' x ' in the | vedclass

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Question

$T_{r+1}={ }^{18} C_{r}\left(q_{x}\right)^{18-r} \cdot\left(\frac{-1}{3 \sqrt{x}}\right)^{r}$

Power of $x$ in $T_{r}+1$ should $6 p O$ $\frac{x^{18

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$T_{r+1}={ }^{18} C_{r}\left(q_{x}ight)^{18-r} \cdot\left(\frac{-1}{3 \sqrt{x}}ight)^{r}$

Power of $x$ in $T_{r}+1$ should $6 p O$ $\frac{x^{18-r}}{x^{1 / 2 r}} \Rightarrow \begin{array}{l}x^{18-r-\frac{1}{2} r} \ \Rightarrow 18-r-\frac{1}{2} r=0\end{array}$

$18-\frac{3}{2} r=0$

$36-3 r=0$

$r=12$

$16 / 129^{6}\left(\frac{-1}{3}ight)^{12}$

$\Rightarrow{ }^{18} C _{12} \frac{9^{1}}{9^{6}} \Rightarrow{ }^{18} C _{12}= T _{13}$

${ }^{18} C_{12}=\alpha \cdot{ }^{18} C_{12}$

$\alpha=1$

The term independent of $' x '$ in the expansion of ${\left( {9\,x\,\, - \,\,\frac{1}{{3\,\sqrt x }}} \right)^{18}}, x > 0$ , is $\alpha$ times the corresponding binomial co-efficient . Then $' \alpha '$ is :

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