Question

Find the 4th term from the end in the expansion of ${(\frac{x^3}{2}-\frac{2}{x^2})}^9$

Answer 1

The general term in the given expansion is given by
$T_{r+1}=(-1{)}^r{\times }^9{C}_r\times {\left(\frac{x^3}{2}\right)}^{9-r}\times {\left(\frac{2}{x^2}\right)}^r$
Now pth term from the end
= $(n-p+2{)}^{th}$ term from the beginning.
$\therefore$ 4th term from the end
= $(9-4+2{)}^{th}$ term from the beginning
= 7th term from the beginning
$=T_7=T_{(6+1)}=(-1{)}^6{\times }^9{C}_6\times {\left(\frac{x^3}{2}\right)}^{(9-6)}\times {\left(\frac{2}{x^2}\right)}^6{=}^9{C}_3\times {\left(\frac{x^3}{2}\right)}^3\times {\left(\frac{2}{x^2}\right)}^6=(\frac{9\times 8\times 7}{3\times 2\times 1}\times \frac{x^9}{8}\times \frac{64}{x^{12}})=\frac{672}{x^3}$
Hence, the 4th term from the end is $\frac{672}{x^3}$