CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

The coefficient of x10 in the expansion of [1+x2(1x)]8 is

A
400
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
476
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
470
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
576
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 476
We have,
[1+x2(1x)]8=(1+x2x3)8=k1+k2+k3=88!k1!k2!k3!(1)k1(x2)k2(x3)k3=k1+k2+k3=88!k1!k2!k3!(1)k3x2k2+3k3

Coefficient of x10, so
2k2+3k3=10k1+k2+k3=8

So,
(1) k1=4,k2=2,k3=2(2) k1=3,k2=5,k3=0

Therefore, the required coefficient
=8!4!2!2!+8!3!5!0!=87654+8766=420+56=476

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon