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Evaluate the sum: S = sumₚ₌₁³² (3p + 2) * [ sum_(q=1)¹⁰ ( sin(2q*pi/11) - i*cos(2q*pi/11)) ]^p.

1. 48 + 2i
2. -48 + 2i
3. 48 - 2i
4. 1648

Correct answer: 48 + 2i

Solution

The inner bracket reduces to -i*(-1) = i, so S = sum (3p+2) i^p; grouping the powers of i in cycles of 4 over p=1..32 yields 48 + 2i.

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