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Q.
Find the $4^{th}$ term from the end in the expansion of $\left(\frac{x^{3}}{2}- \frac{2}{x^{2}}\right)^{9} $.
Binomial Theorem
Solution:
$4^{th}$ term from the end is $9 - 4 + 2$, i.e., $7^{th}$ term from the beginning.
$\therefore T_{7} = \,{}^{9}C_{6}\left(\frac{x^{3}}{2}\right)^{3} \left(\frac{-2}{x^{2}}\right)^{6}$
$= \,{}^{9}C_{3} \frac{x^{9}}{8}\cdot\frac{64}{x^{12}}$
$= \frac{9\times8\times 7}{3\times 2\times 1}\times \frac{8}{x^{3}}$
$= \frac{672}{x^{3}}$