We know that the (r + 1)th term in the expansion of $(x + a)^n$ is given by $t_{r+1}$ = $\binom{n}{r} x^{n-r} a^r$ Then (r + 1)th term in the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$ is given by $T_{r+1}$ = $\binom{18}{r} (9x)^{18-r} \left(-\frac{1}{3\sqrt{x}}\right)^r$ For 13th term, we have r + 1 = 13 i.e., r = 12 ∴ $T_{13}$ = $\binom{18}{12} (9x)^{6} \left(-\frac{1}{3\sqrt{x}}\right)^{12}$ = 18564 $(9x)^6 \frac{1}{3^{12} x^6}$ = 18564 $\frac{9^6}{9^6} \frac{x^6}{x^6}$ = 18564