Question

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

Answer 1

$Q\to (9x-\frac{1}{3\sqrt{2}}{)}^{18}$

So there are $(18+1)=19$ terms

$\therefore 7^{th}$ term from end = ${13}^{th}$ term from fourth

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

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

$\therefore t_{13}=\frac{18!}{12!6!}\times 9^6\times x^6\times \frac{1}{3^{12}}$

$=\frac{13\times 14\times 15\times 16\times 17\times 18}{6\times 5\times 4\times 3\times 2}\times \frac{3^{12}}{3^{12}}$

$=\frac{13\times 7\times 9\times 17}{2}=\frac{51\times 153}{2}$

$=6961.5$