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The expansion of (1 + x + x²)^(3n+1) is written as a₀ + a₁*x + a₂*x² +... + a_(6n+2)*x^(6n+2). Find the value of the sum: S = sum_(r=0)ⁿ [ a_(3r) - (a_(3r+1) + a_(3r+2))/2 ].

1. 0
2. 1
3. 2
4. 3

Correct answer: 0

Solution

Using cube roots of unity, the required sum reduces to [f(1) + f(omega) + f(omega²) evaluated with appropriate weights]/3 type expressions. For (1+x+x²)^(3n+1), substituting omega gives (1+omega+omega²)^(3n+1) = 0, and the sum evaluates to 0.

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