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By considering the geometric series $1+z+z^{2}+...+z^{n-1}$ where $z=\cos(\theta)+i\sin(\theta)$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+...+\cos(n-1)\theta$ = ${1-\cos(\theta)+\cos(n-1)\theta-\cos(n\theta)}\over {2-2\cos(\theta)}$

I've tried expressing $\cos(n\theta)$ as ${e^{in\theta}+e^{-in\theta}} \over {2}$ but I don't think that will lead anywhere. Does it help that $1+z+z^{2}+z^{3}+...+z^{n-1}$=$e^{0i\theta}+e^{i\theta}+e^{2i\theta}+e^{3i\theta}+...+e^{(n-1)i\theta}$?

So the sum $\sum_{r=0}^{n-1} e^{ir\theta}$=${e^{ni\theta}-1} \over {e^{i\theta}-1}$

Thank you in advance :)

Anthony Muleta
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3 Answers3

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Your sum can be rewritten: $\Re(\sum \exp{(i n \theta)})$ which is simply a geometric sum. Then make apparent the real and imaginary parts in your result.

Matt B.
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$$M=cos(\theta)+cos(2\theta)+cos(3\theta) +....+cos((n-1)\theta)=\\\frac{2sin(\frac{\theta}{2})}{2sin(\frac{\theta}{2})}M=\\\frac{2sin(\frac{\theta}{2})}{2sin(\frac{\theta}{2})}cos(\theta)+cos(2\theta)+cos(3\theta) +....+cos((n-1)\theta)=\\ \frac{(sin(\frac{3\theta}{2})-sin(\frac{1\theta}{2}))+(sin(\frac{5\theta}{2})-sin(\frac{3\theta}{2}))+(sin(\frac{7\theta}{2})-sin(\frac{5\theta}{2}))++...+(sin(\frac{(\frac{n+1}{2}\theta}{2})-sin(\frac{n-1}{2}\frac{1\theta}{2}))}{2sin(\frac{\theta}{2})}\\so\\M= \frac{-sin(\frac{1\theta}{2})+sin(\frac{n+1}{2}\theta)}{2sin(\frac{\theta}{2})}\\then\\find \\M+1\\M+1=1+\frac{-sin(\frac{1\theta}{2})+sin(\frac{n+1}{2}\theta)}{2sin(\frac{\theta}{2})} $$

Khosrotash
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Hint: Multiply the numerator and denominator in the sum you computed by the complex conjugate of the denominator (i.e., $e^{-i\theta}-1$).

Harald Hanche-Olsen
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