Question

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

Answer 1

Here general term in the expansion of ${[9x-\frac{1}{3\sqrt{x}}]}^{18}$:

$T_{r+1}={18}_{C_r}(9x{)}^{18-r}{(-\frac{1}{3\sqrt{x}})}^r$ ...(i)

Putting r=12 in (i)

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

$={18}_{C_{12}}{9}^6{x}^6.(-1{)}^{12}.\frac{1}{3^{12}.x^6}$

$={18}_{C_{12}}\frac{9^6}{3^{12}}={18}_{C_{12}}=18564$