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

Find $13^{th}$ term in the expansion of $\big(9x-\large\frac{1}{3\sqrt x}\big)^{18}$

$\begin{array}{1 1}(A)\;18564\\(B)\;18664\\(C)\;18768\\(D)\;18763\end{array} $

1 Answer

Toolbox:
  • $nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-.....+(-1)^nnC_ra^{n-r}b^r+...nC_n(-b)^n$
$13^{th}$ term
$T_{13}=T_{12+1}=18C_{12}(9x)^{18-12}\big(-\large\frac{1}{3\sqrt x}\big)^{12}$
$\Rightarrow 18C_69^6x^6(-1)^{12}.\large\frac{1}{3^{12}}\times \frac{1}{x^6}$
$\Rightarrow 18564\times (3^2)^6.\large\frac{1}{3^{12}}.\frac{x^6}{x^6}$
$\Rightarrow 18564\times \large\frac{3^{12}}{3^{12}}$
$\Rightarrow 18564$
Hence (A) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 
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