Calculate the sum of the series that contain trigonometric funtion $\sum\limits_{k = 0}^n {{2^k}\cos \frac{{k\pi }}{3}} $

Question

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.

Then remove trigonometry: try to compute $\cos(k\pi/3)$ for some values of $k$, then notice a pattern. — Jack D'Aurizio, Feb 02 '17 at 09:11
See http://math.stackexchange.com/questions/2098275/another-way-to-express-textcis-75-circ-textcis-83-circ-textcis — lab bhattacharjee, Feb 02 '17 at 09:34
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 ?! — user90369, Feb 02 '17 at 09:41

score 5 · Answer 1 · answered Feb 02 '17 at 09:16

5

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 $\cos(x+2\pi)=\cos(x)$

answered Feb 02 '17 at 09:16

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Calculate the sum of the series that contain trigonometric funtion $\sum\limits_{k = 0}^n {{2^k}\cos \frac{{k\pi }}{3}} $

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