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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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If the term free from $x$ in the expansion of $(\sqrt x-\large\frac{k}{x^2})^{10}$ is $405$,find the value of $k$.

$\begin{array}{1 1}(A)\;\pm 1\\(B)\;\pm 2\\(C)\;\pm 3\\(D)\;\pm 4\end{array} $

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1 Answer

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Toolbox:
  • $T_{r+1}=nC_ra^{n-r}b^r$
Given :
$(\sqrt x-\large\frac{k}{x^2})^{10}$
$\Rightarrow 10C_r(\sqrt x)^{10-r}(\large\frac{k}{x^2})^r$
$\Rightarrow 10C_r(x)^{\Large\frac{10-r}{2}}.\large\frac{k^r}{x^{2r}}$
$\Rightarrow 10C_r k^r x^{\Large\frac{10-r}{2}-2r}$
Since the term is independent of x we have
$\large\frac{10-r}{2}$$-2r=0$
$10-r-4r=0$
$10-5r=0$
$r=2$
Hence $3^{rd}$ term is independent of $x$
$T_3=10C_2k^2=405$
$\large\frac{10!}{2!8!}$$k^2=405$
$\large\frac{10\times 9\times 8!}{2\times 8!}$$k^2=405$
$45k^2=405$
$k^2=\large\frac{405}{45}$
$k^2=9$
$k=\pm 3$
Hence (C) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

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