A Cube Root of Unity Trick

Problem

Let

(x2+x−1)100=a0+a1x+⋯+a199x199+a200x200.\left(x^2 + x - 1\right)^{100} = a_0 + a_1x + \cdots + a_{199}x^{199} + a_{200}x^{200}.

Find the value of

M=2a0−a1−a2+2a3−a4−a5+⋯+2a198−a199−a200.M = 2a_0 - a_1 - a_2 + 2a_3 - a_4 - a_5 + \cdots + 2a_{198} - a_{199} - a_{200}.

Answer

Solution

Difficulty7/10
Topicscomplex numbers, Roots of Unity Filter, algebra, Binomial Theorem, Polynomials

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