First slide

Binomial theorem for positive integral Index

Question

In the expansion of ${(x^{3} - \frac{1}{x^{2}})}^{15},$ the constant term equals

Moderate

A

$^{15} C_{9}$
B

0
C

$- (^{15} C_{9})$
D

$^{15} C_{11}$

Solution

$T_{r + 1},$ the $(r + 1) t h$ term is the expansion of ${(x^{3} - \frac{1}{x^{2}})}^{15}$ is
$T_{r + 1} =^{15} C_{r} {(x^{3})}^{15 - r} {(- \frac{1}{x^{2}})}^{r} =^{15} C_{r} (- 1)^{r} x^{45 - 5 r}$
To obtain constant term, we set $45 - 5 r = 0 \Rightarrow r = 9$
$∴$ Coefficient of constant term is $^{15} C_{9} (- 1)^{9} = -^{15} C_{9} .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App