# Q6.    Find the 13th term in the expansion of    $\left(9x - \frac{1}{3\sqrt x} \right )^{18},\ x\neq 0$

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $13^{th}$ term of the expansion of     $\left(9x - \frac{1}{3\sqrt x} \right )^{18}$   is

$\\\Rightarrow T_{13}= T_{12+1}\\=^{18}C_{12}(9x)^{18-12}\left(\frac{1}{3\sqrt{x}}\right)^{12}\\ \\ \\=\frac{18!}{12!6!}\times9^{6}\left ( \frac{1}{3} \right )^{12}\times x^{6-6}$

$\\=\frac{18\times17\times16\times15\times14\times13}{6\times5\times4\times3\times2}\times9^{6}\left ( \frac{1}{3^{12}} \right )$

$=18564$

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