Given:
x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n}
Required: Find: \sum_{k=1}^{n} x_k
This is the sum of all n^\text{th} roots of unity (from k = 1 to n ).
We know: \sum_{k=0}^{n-1} e^{2\pi i k/n} = 0 So shifting index from k = 1 to n just cycles the same roots: \sum_{k=1}^{n} e^{2\pi i k/n} = 0
✅ Final Answer: \boxed{0}
A particle P starts from the point
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