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This is the series: $$\sum\limits_{k = 0}^n {{2^k}\cos \frac{{k\pi }}{3}} $$

Because the series contain trigonometry so that I've tried a lot but still can't solve it with common methods. Many thanks to all that can solve it for me. This question is in my seminar test at university.

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    hint: $\cos(kx) = \dfrac{e^{ikx}+ e^{-ikx}}{2}$ – 2017-02-02
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    Then remove trigonometry: try to compute $\cos(k\pi/3)$ for some values of $k$, then notice a pattern. – 2017-02-02
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    [This](http://www.wolframalpha.com/input/?i=Table%5BCos%5Bk*Pi%2F3%5D,%7Bk,0,11%7D%5D) might give you an idea ! – 2017-02-02
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    See http://math.stackexchange.com/questions/2098275/another-way-to-express-textcis-75-circ-textcis-83-circ-textcis – 2017-02-02
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    The result is $\displaystyle\frac{2^{n+1}}{\sqrt{3}}\sin\frac{(n+1)\pi}{3}$ . You've got several senseful proposals - do you see how you can get the solution ?! – 2017-02-02

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Hint:

$$\cos\frac{1\pi}{3}=\frac{1}{2}\\ \cos\frac{2\pi}{3}=-\frac{1}{2}\\ \cos\frac{3\pi}{3}=-1\\ \cos\frac{4\pi}{3}=-\frac{1}{2}\\ \cos\frac{5\pi}{3}=\frac{1}{2}\\ \cos\frac{6\pi}{3}=1 $$

For $k>6$, use the fact that $\cos(x+2\pi)=\cos(x)$