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Let (1+x+x2)2014=a0+a1x+a2x2+a3x3+...+a4028x4028, and let A=a0a3+a6...+a4026,B=a1a4+a7...a4027,C=a2a5+a8...+a4028.
Then

A
|A|=|B|>|C|
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B
|A|=|B|<|C|
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C
|A|=|C|>|B|
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D
|A|=|C|<|B|
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Solution

The correct option is D |A|=|C|<|B|
(1+x+x2)2014=a0+a1x+a2x2+a3x3+...+a4028x4028
Put x=1
1=a0a1+a2a3+a4a5+a6.....+a4028....(1)
Put x=ω
(1ω+ω2)2014=(2ω)2014=a0a1ω+a2ω2a3+...+a4028ω4028....(2)
Put x=ω2
(1ω2+ω4)2014=(2ω2)2014=a0a1ω2+a2ωa3+...+a4028ω9056....(3)
Add equation (1)+(2)+(3)
1+(2ω)2014+(2ω2)2014=3(a0a3+a6......)
A=a0a3+a6.....=1+22014ω+22014ω23
A=1220143|A|=2201413
Now, equation (1)+(2)×ω+(3)×ω2
1+22014ω2014.ω+22014(ω2)2014.(ω2)3=a2a5+a8+.......
1+22014(ω2015+ω4030)3=a2a5+a8+.......=C
C=1220143|C|=2201413
and similarly equation (1)+(2)×ω2+(3)×ω
B=1+22014.(ω2016+ω4029)3
|B|=1+220153
|B|>|A|=|C|

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