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If the sum of the coefficients of $1^{st},\:2^{nd}$ and $3^{rd}$ terms in the expansion of $(x^2+\large\frac{1}{x})$$^n $ is $46$, then the coefficient of the term independent of $x$ = ?

$\begin{array}{1 1} 106 \\ 92 \\ 98 \\ 84 \end{array} $

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Given $^nC_0+^nC_1+^nC_2=46$
$\Rightarrow\:1+n+\large\frac{n(n-1)}{2!}$$=46$
$\Rightarrow\:n^2+n-90=0$ or $n=9$ since $n>0$
General term in the expansion $(x^2+\large\frac{1}{x}$$)^n$ is
$^nC_r.\large(x^2)^{n-r}.(\frac{1}{x})^r$
$=^nC_r.x^{2n-3r}$
For the term independent of $x$, $2n-3r=0$
$\Rightarrow\:r=\large\frac{2n}{3}$$=6$
$\therefore$ The term independent of $x$ = $^9C_6=84$
answered Oct 20, 2013 by rvidyagovindarajan_1
 

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