Sail E0 Webinar
Question
(2+x+x3)5=a0+a1x+.....+a15x15. find the value of a0+2a1+a2+2a3....+2a15 
Options:
A .  1024
B .  1536
C .  1586
D .  2048
E .  1200 m
Answer: Option B
:
B
At x=1
a0+a1+.....+a15=45.............(1)
At x= -1
a0a1+......a15=0.................(2)
(1)-(2)=2a1+2a3.......2a15=45.........(3)
([(1)+(2)]2)=a0+a2...=452........(4)
(3)+(4)=a0+2a1+a2+2a3....+2a15=45+[452]=1536.

Was this answer helpful ?
Next Question

Submit Solution

Your email address will not be published. Required fields are marked *

More Questions on This Topic :


Latest Videos

Latest Test Papers