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Consider the expression $$(1 + x + x^2)^n = A_0 + A_1x + A_2x^2+\cdots + A_{2n-1}x^{2n-1} + A_{2n}x^{2n}$$ (where $n$ belongs to positive integers).

$n\equiv 0\mod 4 \implies \quad A_0 - A_2 + A_4 - A_6 + A_{8}+\cdots = 1$

$n \equiv 2 \mod 4 \implies \quad A_0 - A_2 + A_4 - A_6 + A_{8}+\cdots = -1$

I know it can be directly obtained by putting x=i, but I was interested in getting the results by remaining in the real domain. Thanks for being so helpful.

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