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If the $4^{th}$ term in the expansion of $\big(x^2-\large\frac{1}{x^3}\big)^n$ is independent of $x$, then $n=?$

$\begin{array}{1 1} 4 \\ 5 \\ 6 \\ None\;of\;the\;above \end{array} $

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$T_4$ in the expansion of $\big(x^2-\large\frac{1}{x^3}\big)$ is
$^nC_3.(x^2)^{n-3}.\big(\large\frac{1}{x^3}\big)^3$
$=^nC_3.x^{2n-15}$
For independent term $2n-15=0$
$\Rightarrow\:n=\large\frac{15}{2}$ which is not possible.
answered Oct 26, 2013 by rvidyagovindarajan_1
 

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