Question

# Let (1+x+x2)2014=a0+a1x+a2x2+a3x3+...+a4028x4028, and let A=a0−a3+a6−...+a4026,B=a1−a4+a7−...−a4027,C=a2−a5+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=a0−a1+a2−a3+a4−a5+a6−.....+a4028....(1) Put x=−ω (1−ω+ω2)2014=(−2ω)2014=a0−a1ω+a2ω2−a3+...+a4028ω4028....(2) Put x=−ω2 (1−ω2+ω4)2014=(−2ω2)2014=a0−a1ω2+a2ω−a3+...+a4028ω9056....(3) Add equation (1)+(2)+(3) ⇒1+(2ω)2014+(2ω2)2014=3(a0−a3+a6......) ⇒A=a0−a3+a6.....=1+22014ω+22014ω23 A=1−220143⇒|A|=22014−13 Now, equation (1)+(2)×ω+(3)×ω2 ⇒1+22014ω2014.ω+22014(ω2)2014.(ω2)3=a2−a5+a8+....... ⇒1+22014(ω2015+ω4030)3=a2−a5+a8+.......=C C=1−220143⇒|C|=22014−13 and similarly equation (1)+(2)×ω2+(3)×ω B=1+22014.(ω2016+ω4029)3 |B|=1+220153 ∴|B|>|A|=|C|

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