First slide

Binomial theorem for positive integral Index

Question

$(p + 2)$ th term from the end in the binomial expansion of
${(x^{2} - \frac{2}{x^{2}})}^{2 n + 1}$ is

Moderate

A

$^{2 n + 1} C_{2 n - n} (- 2)^{2 n - p} x^{2 p + 1 - 2 n}$
B

$^{2 n + 1} C_{2 n - n} (- 2)^{2 n - p} x^{2 n - 2 p}$
C

$^{2 n + 1} C_{2 n - p} (- 2)^{2 n - p} x^{2 n - 2 p + 1}$
D

$^{2 n + 1} C_{2 n - p} (- 2)^{2 n - p} x^{2 p - 1 + 2 n}$

Solution

$(p + 2)$ th term from the end
$= [2 n + 2 - (p + 1)] th = (2 n - p + 1) th$
from the beginning
and $T_{2 n - p + 1} =^{2 n + 1} C_{2 n - p} {(x^{2})}^{2 n + 1 - (2 n - p)} {(- \frac{2}{x^{2}})}^{2 n - p}$
$=^{2 n + 1} C_{2 n - p} (- 2)^{2 n - p} x^{2 p + 1 - 2 n}$

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