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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-A. `|A|=|B|gt|C|`B. `|A|=|B|lt|C|`C. `|A|=|C|gt|B|`D. `|A|=|C|lt|B|` |
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Answer» Correct Answer - D `(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|` |
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