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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The coefficient of the third term in the expansion of $(x^2-\large\frac{1}{4})^n$,when expanded in descending powers of $x$ is $31$.Then n is equal to

$\begin{array}{1 1}(A)\;16\\(B)\;30\\(C)\;32\\(D)\;\text{None of these}\end{array} $

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1 Answer

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  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_3=nC_2 (x^2)^{n-2} (-\large\frac{1}{4})^2$
$\quad=\large\frac{n(n-1)}{32}$$ x^{2n-4}$
Coefficient of $T_3=\large\frac{n(n-1)}{32}$
$\therefore \large\frac{n(n-1)}{32}=$$31$
$\Rightarrow n(n-1)=32(32-1)$
$\Rightarrow n=32$
Hence (C) is the correct answer.
answered Jun 25, 2014 by sreemathi.v
 

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