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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The coefficient of $x^2$ in the expansion of $(1+4x+x^2)^{\Large\frac{1}{2}}$ is

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

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1 Answer

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  • $T_{r+1}=nC_r a^{n-r} b^r$
$(1+4x+x^2)^{\Large\frac{1}{2}}=1+\large\frac{1}{2}$$(4x+x^2)+\large\frac{\Large\frac{1}{2}(-\frac{1}{2})}{1.2}$$(4x+x^2)+$......
$\Rightarrow 1+2x+\large\frac{x^2}{2}-\frac{1}{8}$$(16x^2+x^4+8x^3)+$.......
$\therefore$ Coefficient of $x^2=\large\frac{1}{2}$$-2$
$\Rightarrow -\large\frac{3}{2}$
Hence (D) is the correct answer.
answered Jun 26, 2014 by sreemathi.v
 
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