CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If a=nC0+nC3+nC6+...
b=nC1+nC2+nC4+nC5+nC7+nC8+...
c=nC1nC2+nC4nC5+nC7nC8+...,
then the value of (ab2)2+3c24 is

Open in App
Solution

Let ω be the cube root of unity.
(1+ω)n=nC0+nC1ω+nC2ω2+....
Since, ω3n=1, ω3n+1=ω, ω3n+2=ω2, n=0,1,2,...
(1+ω)n=(nC0+nC3+...)+(nC1+nC4+...)(1+3i2)+(nC2+nC5+...)(13i2)
=(nC0+nC3+...)12(nC1+nC2+nC4+nC5+....)+i32(nC1nC2+nC4nC5+....)

(1+ω)n=(ab2)+i3c2
Taking modulus both the side, we get
|(ω2)n|=(ab2)2+3c24 [1+ω+ω2=0]
(ab2)2+3c24=1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Domestic Electric Circuits
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon