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## Homework Statement

Prove that for any [itex]n \in \mathbb{N}[/itex] and [itex]x \in \mathbb{R}[/itex], we have

[tex]\sum_{k = 0}^{n} {\cos{(kx)}} = \frac{1}{2}+ \frac{\cos{(nx)} - \cos{[(n+1)x]}}{2 - 2\cos {x}}[/tex]

## Homework Equations

None I can think of.

## The Attempt at a Solution

Try induction. The result holds if n = 0. Suppose the result holds for some natural number n. Then we get

[tex]\sum_{k = 0}^{n + 1} {\cos{(kx)}} = \sum_{k = 0}^{n} {\cos{(kx)} + \cos{[(n + 1)x]}} = \frac{1}{2}+ \frac{\cos{(nx)} - \cos{[(n+1)x]}}{2 - 2\cos {x}} + \cos{[(n + 1)x]}[/tex]

Now I could collect the denominators but it doesn't help (as far as I can see). I'm stuck at this point. Please help me out!!

For some perspective ... This is a third year complex analysis course and this is my first assignment. My first year algebra course covered induction and complex numbers, so I can't tell if the purpose of this question is review or if there's a deeper trick to it. The rest of the problems on this assignment are on the topology of the complex plane so this problem is kind of a sore thumb. It seems like it should be "easier" but I'm having a hard time. :(

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