Question

Fidn the $7^{th}$ term from the end in the expansion of ${(9x-\frac{1}{3\sqrt{x}})}^{18},x\ne 0$.

Answer 1

${(9x-\frac{1}{3\sqrt{x}})}^{18}=7^{th}$ term from end

${11}^{th}$ term from start

$T_{11}={}^{18}{C}_{11}(9x{)}^{11}{\left(\frac{1}{3\sqrt{x}}\right)}^{18-11}$.