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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The co-efficient of $x$ in the expansion of $(x^2+\large\frac{\lambda}{x})^2$ is

$\begin{array}{1 1}(A)\;20\lambda\\(B)\;10\lambda\\(C)\;10\lambda^3\\(D)\;20\lambda ^3\end{array} $

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1 Answer

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Toolbox:
  • $T_{r+1}=nC_ra^{n-r} b^r$
$T_{r+1}=5C_r (x^2)^{5-r} (\large\frac{\lambda}{x})^r$
$\Rightarrow 5C_r\lambda ^r x^{10-3r}$
Let $T_{r+1}$ contains $x$
$\therefore 10-3r=1$
$-3r=1-10$
$-3r=-9$
$r=3$
$\therefore T_{r+1}=T_{3+1}$
$\Rightarrow 5C_3 \lambda ^3 x^{10-3(3)}$
$\Rightarrow 10\lambda ^3x$
$\therefore$ Coefficient of x=$10\lambda ^3$
Hence (C) is the correct answer.
answered Jun 25, 2014 by sreemathi.v
 

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