Question

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

• A. $\frac{674}{x}$
• B. $\frac{670}{x^3}$
• C. $\frac{672}{x^3}$
• D. $\frac{-670}{x^3}$

Answer 1

Answer: (C)

$4^{th}$ term from the end is $9-4+2$, i.e., $7^{th}$ term from the beginning.
$\therefore T_7={}^9{C}_6{\left(\frac{x^3}{2}\right)}^3{\left(\frac{-2}{x^2}\right)}^6$
$={}^9{C}_3\frac{x^9}{8}\cdot \frac{64}{x^{12}}$
$=\frac{9\times 8\times 7}{3\times 2\times 1}\times \frac{8}{x^3}$
$=\frac{672}{x^3}$