# Stuck at summation of exponential.

I want to find the coefficient but I am stuck. The formula is:

$$\frac{1}{10}\sum_{n=0}^{9} e^{-jk\omega_0n}$$

In order to solve this I am using the following formula:

$$\sum_{n_1}^{n_2} \alpha^{n} = \frac{\alpha^{n_1}-\alpha^{n_2+1}}{1-\alpha}$$ I have tried everything but I am unable to reach the answer. The answer is: $$\frac{1}{10}(1-e^{-j\omega_08k})$$

$$\displaystyle S=\cfrac 1{10}\sum_{n=0}^9 e^{-jk\omega_0 n}=\cfrac 1{10}\cdot\cfrac{1-e^{-j10k\omega_0}}{1-e^{-jk\omega_0}}=\cfrac 1{10}\cdot \left(1+\cfrac{e^{-jk\omega_0}(1-e^{-j9k\omega_0})}{1-e^{-jk\omega_0}}\right)$$

And $$1-e^{j\theta}=1-\cos\theta-j\sin\theta=2\sin\cfrac{\theta}{2}\left(\sin\cfrac{\theta}{2}-j\cos\cfrac{\theta}{2}\right)=2\sin\cfrac{\theta}{2}e^{j(\theta-\pi)/2}$$

Hence,

$$\cfrac{e^{-jk\omega_0}(1-e^{-j9k\omega_0})}{1-e^{-jk\omega_0}}= \cfrac {e^{-jk\omega_0}2\sin (-9k\omega_0/2) e^{-j4k\omega_0}}{2\sin(-k\omega_0/2)}$$

$$=\cfrac {e^{-j5k\omega_0}\sin (9k\omega_0/2) }{\sin(k\omega_0/2)}$$

$$S=\cfrac 1{10}\left(1+\cfrac {e^{-j5k\omega_0}\sin (9k\omega_0/2) }{\sin(k\omega_0/2)}\right)$$

It seems the definition of $$\omega_0$$ is needed if we want to go further.

====================

Edit: with integer $$k$$, generally we have

$$\displaystyle \sum_{n=0}^{N-1}e^{-j2kn\pi/N}=\cfrac {1-e^{-j2k\pi}}{1-e^{-j2k\pi/N}}=0$$

The correct answer you provided doesn't evaluate to $$0$$ with, say, $$k=1$$. So something must be wrong.

• $w_0$ is $(2\pi)/10$ – Ahmad Qayyum Nov 8 '18 at 17:50
• can you complete the solution? – Ahmad Qayyum Nov 8 '18 at 18:26
• @Ahmad Qayyum. If $k$ is integer, the sum apparently evalutes to $0$, as $\cfrac {1-e^{-j2\pi k}}{1-e^{-jk2\pi/10}}$ is $0$. Take $k=1$, your answer is $\cfrac {1-e^{-j8/5\pi}}{10}$ which is not $0$. Something still not right. – Lance Nov 8 '18 at 18:34