Write down the sum of the geometric series $z+z^2+\cdots+z^n$, where $n$ is a positive inte

Question

Write down the sum of the geometric series $z+z^2+\cdots+z^n$, where $n$ is a positive integer and $z\ne1$.[1] By putting $z=\mathrm{e}^{i\theta}$, where $\theta\in\mathbb{R}$ and $\sin(\theta/2)\ne0$, in your result, show that this sum can be written as $\dfrac{\sin(n\theta/2)}{\sin(\theta/2)}\mathrm{e}^{i(n+1)\theta/2}$.[2] Hence, by using the identity $\mathrm{e}^{i\theta}=\cos\theta+i\sin\theta$, or otherwise, show that $\sin\theta+\sin2\theta+\cdots+\sin n\theta=\dfrac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}$.[3]

Accepted Answer

(c)(i) $\dfrac{z(1-z^n)}{1-z}$ (ii) $\dfrac{\sin(n\theta/2)}{\sin(\theta/2)}\mathrm{e}^{i(n+1)\theta/2}$ (iii) $\displaystyle\sum_{r=1}^{n}\sin(r\theta)=\dfrac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}$.

2021 NYJC P2 Q3

2021 NYJC P2 Q3

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