Question

Find the ${13}^{th}$ terms in the expansion of ${(9x-\frac{1}{3\sqrt{x}})}^{18},x\ne 0$.

• A. $18564$
• B. $87328$
• C. $17374$
• D. $35546$

Answer 1

The correct option is A $18564$

${\left(9x-\frac{1}{\sqrt{3x}}\right)}^{18}$

$T_{r+1}={}^{18}{C}_r{\left(9x\right)}^{18-r}{\left(-\frac{1}{3\sqrt{x}}\right)}^r$

putting $r=12$

$T_{13}={}^{18}{C}_{12}{\left(9x\right)}^{18-12}{\left(-\frac{1}{3\sqrt{x}}\right)}^r$

$={}^{18}{C}_6\times {\left(9x\right)}^6\times \frac{1}{3^{12}\times x^6}$

$={}^{18}{C}_6\times 9^6\times x^6\times \frac{1}{3^{12}\times x^6}$

$={}^{18}{C}_6\times 3^{12}\times \frac{1}{3^{12}}$

$=\frac{18\times 17\times 16\times 15\times 14\times 13}{6\times 5\times 4\times 3\times 2}$

$=17\times 4\times 3\times 7\times 13$

$=18564$