Question

$\text{Let}(1+x+x^2{)}^{2014}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+....+a_{4028}{x}^{4028}\text{and let}A=a_0-a_3+a_6-......+a_{4026,}B=a_1-a_4+a_7-......-a_{4027,}C=a_2-a_5+a_8-......+a_{4028}\text{Then}$

• A. |A|=|B|>|C|
• B. |A|=|B|<|C|
• C. |A|=|C|>|B|
• D. |A|=|C|<|B|

Answer 1

The correct option is C

|A|=|C|>|B|

$(1+x+x^2{)}^{2014}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6+.....\text{Substituting}-1,-\omega ,-{\omega }^2\text{where}-\omega =e^{i2\frac{\pi }{3}}1=a_0-a_1+a_2-a_3+a_4-a_5+a_6...(i\left)\right(1-\omega +{\omega }^2{)}^{2014}=a_0-a_1\omega +a_2{\omega }_2-a_3+a_4\omega -a_5{\omega }^2+a_6.........\left(ii\right)(1-{\omega }^2+\omega {)}^{2014}=a_0-a_1{\omega }^2+a_2\omega -a_3+a_4{\omega }^2-a_5\omega +a_6.......(iii\left)\right(i)+(ii)+(iii)\Rightarrow a_0-a_3+a_6....=\frac{1+2^{2014}(\omega +{\omega }^2)}{3}=\frac{1-2^{2014}}{3}(i)+{\omega }^2(ii)+\omega (iii)\Rightarrow a_1-a_4+a,....=-\left(\frac{1+2^{2015}}{3}\right)(i)+\omega (ii)+{\omega }^2(iii)\Rightarrow a_2-a_5+a_8....=\frac{1-2^{2014}}{3}$
$\Rightarrow$ |A|=|C|<|B|