Browse Questions

# 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

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

Toolbox:
• $T_{r+1}=nC_r a^{n-r} b^r$
$T_3=nC_2 (x^2)^{n-2} (-\large\frac{1}{4})^2$
$\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.