Question

In the expansion of ${(3x-\frac{1}{x^2})}^{10}$, the $5^{th}$ term from the end is

• A. $\frac{16486}{x^8}$
• B. $\frac{17010}{x^8}$
• C. $\frac{13486}{x^8}$
• D. None of these

Answer 1

The correct option is B $\frac{17010}{x^8}$

There are $11$ terms in the expansion of ${(3x-\frac{1}{x^2})}^{10}$
Therefore, $5^{th}$ term from the end.
$=(10-5+2{)}^{th}$ term from beginning
$=T_7={{}^{10}C}_6{\left(3x\right)}^{10-6}{(-\frac{1}{x^2})}^6$
$=\frac{10!}{6!4!}{3}^4{x}^4\frac{1}{x^{12}}$

$=\frac{17010}{x^8}$