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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

In the expansion of $(x^2-\large\frac{1}{x^2})^{16}$ the value of the constant term is ________

$\begin{array}{1 1}(A)\;16C_{16}\\(B)\;16C_4\\(C)\;16C_9\\(D)\;16C_8\end{array} $

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1 Answer

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  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_{r+1}=16C_r (x^2)^{16-r} (-\large\frac{1}{x^2})^r$
$\Rightarrow 16C_r (-1)^r (x^2)^{16-r}.(-\large\frac{1}{x^2})$
$\Rightarrow (-1)^r 16C_r x^{32-2r-2r}$
$\Rightarrow (-1)^r 16C_r x^{32-4r}$
Let $T_{r+1}$ be the constant term
$32-4r=0$
$32=4r$
$r=8$
$T_{8+1}=16C_8 (-1)^8$
$\Rightarrow 16C_8$
Hence (D) is the correct answer.
answered Jun 25, 2014 by sreemathi.v
 

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