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Tried to prove this by writing out as I hoped the e terms would simplify, but it wasn't the case so I am stuck now. Am I missing a summation property I could use?

$$\sum_{n=0}^{n=5}(\frac{1}{2}e^{\frac{-i2\pi k}{6}})^n = \frac{1-(\frac{1}{2}e^{\frac{-i2\pi k}{6}})^6 }{1-\frac{1}{2}e^{\frac{-i2\pi k}{6}} } $$

Hints and/or solutions are appreciated!

dsisko
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1 Answers1

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This is a geometric sum. $$S_n =1+r+ r^2+\cdots + r^n$$

$$rS_n =r+ r^2+ r^3+\cdots + r^{n+1}$$ Subtracting, most terms cancel and you get $$(1-r)S_n=(1+r+ r^2+\cdots + r^n)-(r+ r^2+ r^3+\cdots + r^{n+1})$$ $$(1-r)S_n=1-r^{n+1}$$ If $r\neq 1$, then $$S_n=\frac{1-r^{n+1}}{1-r}$$

Is this what you want to show?

MPW
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