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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

If the $r^{th}$ term in the expansion of $(\large\frac{x}{3}-\frac{2}{x^2})^{10}$ contains $x^4$ then $r$ is equal to

$\begin{array}{1 1}(A)\;4\\(B)\;3\\(C)\;2\\(D)\;\text{none of these}\end{array} $

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1 Answer

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  • $T_{r+1}=nC_ra^{n-r} b^r$
$T_r=T_{(r-1)}+1=10C_{r-1}(\large\frac{x}{3})^{10-(r-1)}(-\large\frac{2}{x^2})^{r-1}$
$\Rightarrow 10C_{r-1}(\large\frac{1}{3})^{n-r} $$(-2)^{r-1} x^{13-3r}$
Let $T_{r+1}$ contains $x^4$
$\therefore 13-3r=4$
(i.e) $r=3$
Hence (B) is the correct answer.
answered Jun 26, 2014 by sreemathi.v
 

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