wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The coefficient of x7 in the expansion of (1xx2+x3)6 is

A
132
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
144
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
132
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
144
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 144
Coff. of x7 in exp. of (1xx2+x3)6 is

(1xx2+x3)6=[(1x)x2(1x)]6

(1xx2+x3)6=(1x2)6(1x)6

(16x2+6C2x466C3x6+6C4x86C5x10+x12)(16x+6C2x26C3x3+6C4x46x5+x6)

Coff. of x7 is (6)(6C3)+6C2(6C3)+(6)(6)

6×2015×20+36

Coff. (x2)=144

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
What is Binomial Expansion?
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon