First slide

Binomial theorem for positive integral Index

Question

if ${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$ then the expression $a_{2} + a_{4} + a_{6} + \dots + a_{12}$ has the value

Moderate

A

32
B

63
C

64
D

None of these

Solution

Given that, ${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$ on putting X=1 and-1 adding the results we get
$64 = 2 (1 + a_{2} + a_{4} + \dots + a_{12})$
$∴ a_{2} + a_{4} + a_{6} + \dots + a_{12} = 31$

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