Let
`(1+x+x^(2))^(2014)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+ . . . +a_(4028)x^(4028),` and let
`A=a_(0)-a^(3)+a_(6)- . . . +a_(4026)`
`B=a_(1)-a_(4)+a_(7)- . . . . . . . -a_(4027)`,
`C=a_(2)-a_(5)+a_(8)- . . . .+a_(4028)`,
Then-

Options

A. `|A|=|B|gt|C|`
B. `|A|=|B|lt|C|`
C. `|A|=|C|gt|B|`
D. `|A|=|C|lt|B|`

Answer

D

Explanation

`(1+x+x^(2))^(2014)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+ . . . +a_(4028)x^(4028),`
put x=-1
`1=1=a_(0)-a_(2)+a_(2)-a_(3)+a_(4)-a_(5)+a_(6)-a_(7). . . .` . . . (1)
`"put"x=-omega`
`(2omega)^(2014)=(1-omega+omega^(2))^(2014)=a_(0)-a_(1omega)+a_(2omega^(2))+a_(3)+a_(4)omega^(2)+a_(6). . . . . . . ` . . .(2)
`"Put"x=-omega^(2)`
`(2omega^(2))^(2014)=(a-omega^(2)+omega)^(2014)=a_(0)-a_(3)+a_(4)omega^(2)+a_(2)omega-a_(3)+a_(4)omega^(2)-a_(5)omega . . . . . . .` (3)
Now, (1)+(2)+(3)
`rArr1+(2omega)^(2014)+(2omega^(2))^(2014)=3(a_(0)-a_(3)+a_(6) . . . . . . . . .`
`rArra_(0)-a_(3)+a_(6). . . . . . . .=(1+2^(2014)omega+2^(2014)omega^(2))/(3)`
`A=(1-2^(2014))/(3)`
`|A|=(2^(2014)-1)/(3)`
`and (1)+(2)xxomega+(3)omega^(2)`
`rArr(1+2^(2014).omega^(2014).omega^(2015)+2^(2014).omega^(4030))/(3)=a_(2)-a_(5)+a_(8). . . . .`
`rArr(1+2^(2014)+omega^(2015)+2^(2014).omega^(4030))/(3)=C`
`rArrC=(1-2^(2014))/(3)rArr|C|=(2^(2014)-1)/(3)`
and similarly `(1)+(2)xxomega^(2)+(3)xxomega`
`B=(1+2^(2014).omega^(2014).omega^(2)+2^(2014).(omega^(2))^(2014).omega)/(3)`
`=(1+2^(2014).omega^(2016)+2^(2014).omega^(4029))/(3)`
`|B|=(1+2^(2015))/(3)`
`:.|B|gt|A|=|C|`