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

If for z as real or complex, (1+z2+z4)8=C0+C1z2+C2z4++C16z32, then

A
C0C1+C2C3++C16=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
C0+C3+C6+C9+C12+C15=37
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
C2+C5+C8+C11+C14=36
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
C1+C4+C7+C10+C13+C16=37
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D C1+C4+C7+C10+C13+C16=37
(1+z2+z4)8=C0+C1z2+C2z4++C16z32 [1]

Putting z=i, where i=1,
(11+1)8=C0C1+C2C3++C16
or, C0C1+C2C3++C16=1

Also, putting z=ω, where ω is cube root of unity
(1+ω2+ω4)8=C0+C1ω2+C2ω4++C16ω32
C0+C1ω2+C2ω++C16ω2=0 [2]

Putting z=ω2,
(1+ω4+ω8)8=C0+C1ω4+C2ω8++C16ω64
C0+C1ω+C2ω2++C16ω=0 [3]

Putting z=1,
38=C0+C1+C2++C16 [4]

Adding [2], [3], [4], we get
3(C0+C3+C6++C15)=38
C0+C3+C6++C15=37

Similarly, first multiplying [1] by z and then putting 1, ω, ω2 and adding, we get
C1+C4+C7+C10+C13+C16=37

Multiplying [1] by z2 and then putting 1, ω, ω2 and adding, we get
C2+C5+C8+C11+C14=37

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