Let (1+x+x^2)^{2014} = a_0 + a_1 x + a_2 x^2 | vedclass

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(D) Let $f(x) = (1+x+x^2)^{2014} = \sum_{r=0}^{4028} a_r x^r$. Let $\omega$ be the complex cube root of unity,such that $1+\omega+\omega^2 = 0$ and $\

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$|A| = |C| < |B|$

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(D) Let $f(x) = (1+x+x^2)^{2014} = \sum_{r=0}^{4028} a_r x^r$. Let $\omega$ be the complex cube root of unity,such that $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. Consider the sum $S_k = \sum_{n \equiv k \pmod 3} a_n$. Using the roots of unity filter,we have: $A = a_0 - a_3 + a_6 - \ldots = \frac{1}{3} [f(i) + f(-i) + f(1)]$ is not applicable here; instead,we use $\omega$. Let $f(x) = (1+x+x^2)^{2014}$. $f(1) = 3^{2014} = a_0 + a_1 + a_2 + \ldots + a_{4028}$. $f(\omega) = (1+\omega+\omega^2)^{2014} = 0^{2014} = 0 = a_0 + a_1 \omega + a_2 \omega^2 + a_3 + a_4 \omega + a_5 \omega^2 + \ldots$. $f(\omega^2) = (1+\omega^2+\omega^4)^{2014} = (1+\omega^2+\omega)^{2014} = 0^{2014} = 0 = a_0 + a_1 \omega^2 + a_2 \omega + a_3 + a_4 \omega^2 + a_5 \omega + \ldots$. By solving these linear equations,we find that $A=B=C$ is not necessarily true. Given the structure $A, B, C$,we evaluate $f(x) = (1+x+x^2)^{2014}$. For $x = -1$,$f(-1) = (1-1+1)^{2014} = 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{4028}$. Using the property of roots of unity,it can be shown that $|A| = |C| < |B|$.

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

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