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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})$$^n is 46, then the coefficient of the term independent of x = ? \begin{array}{1 1} 106 \\ 92 \\ 98 \\ 84 \end{array} Can you answer this question? ## 1 Answer 0 votes Given ^nC_0+^nC_1+^nC_2=46 \Rightarrow\:1+n+\large\frac{n(n-1)}{2!}$$=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 $x$ = $^9C_6=84$