Q. $The\ coefficient\ of\ x^{- 10}\ in\ \left( x^{2} - \frac{1}{x^{3}} \right)^{10}\ is$

A) $- 252$
B) 210
C) $- (5!)$
D) $- 120$
Explanation

(b)
$General\ term\ ,\ T_{r + 1} = \ ^{10}C_{r}(x^{2})^{10 - r}\left( - \frac{1}{x^{3}} \right)^{r}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ ^{10}C_{r}x^{20 - 5r}{( - 1)}^{r}$
Since, this term condition $x^{- 10}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 20 - 5r = - 10 \Longrightarrow r = 6$
$\therefore$ Coefficient of $x^{- 10} = \ ^{10}C_{6}{( - 1)}^{6} = 210$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.