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CIE 9231 2024 June Paper 23 Q6

A Level / CIE / FP2

CIE 9231 2024 June Paper 23 Paper · Question 6

题目

Problem

(a) Show that, for z2≠z−2z^2\ne z^{-2},

∑r=1nz4r=z4n+2−z2z2−z−2.\sum_{r=1}^{n}z^{4r}=\frac{z^{4n+2}-z^2}{z^2-z^{-2}}.
[2]

(b) By letting z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta, show that, if sin⁡2θ≠0\sin2\theta\ne0,

∑r=1nsin⁡(4rθ)=cos⁡2θ−cos⁡(4n+2)θ2sin⁡2θ.\sum_{r=1}^{n}\sin(4r\theta)=\frac{\cos2\theta-\cos(4n+2)\theta}{2\sin2\theta}.
[5]
题目中文翻译

(a) 证明当 z2≠z−2z^2\ne z^{-2} 时，

∑r=1nz4r=z4n+2−z2z2−z−2\sum_{r=1}^{n}z^{4r}=\frac{z^{4n+2}-z^2}{z^2-z^{-2}}

(b) 令 z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta，证明当 sin⁡2θ≠0\sin2\theta\ne0 时，

∑r=1nsin⁡(4rθ)=cos⁡2θ−cos⁡(4n+2)θ2sin⁡2θ\sum_{r=1}^{n}\sin(4r\theta)=\frac{\cos2\theta-\cos(4n+2)\theta}{2\sin2\theta}

解答