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# Coefficient of $x^6$ in the expansion $(x+\large\frac{1}{x^2})^6$ is

$\begin{array}{1 1}(A)\;10\\(B)\;15\\(C)\;16\\(D)\;\text{None of these}\end{array}$

Can you answer this question?

Toolbox:
• $T_{r+1}=nC_r a^{n-r}b^r$
$T_{r+1}=6C_r x^{6-r} (\large\frac{1}{x^2})^r$
$\Rightarrow 6C_rx^{6-3r}$
$6-3r=6$
$\Rightarrow r=0$
$\therefore T_{r+1}=T_{0+1}$
$\Rightarrow 6C_0 x^6$
$\Rightarrow x^6$
$\therefore$ Coefficient of $x^6=1$
Hence (D) is the correct answer.
answered Jun 26, 2014