Answer: Proving Sums with De Moivre's Theorem

[rock.freak667]

Using De Moivre's theorem to prove the sum of a series

Homework Statement

Write down an expression in terms of z and N for the sum of the series:
$$\sum_{n=1}^N 2^{-n} z^n$$

Use De Moivre's theorem to deduce that

$$\sum_{n=1}^{10} 2^{-n} \sin(\frac{1}{10}n\pi)$$ = $$\frac{1025\sin(\frac{1}{10}\pi)}{2560-2048\cos(\frac{1}{10}\pi)}$$

Homework Equations

$$e^{in\theta}=(\cos{\theta}+i\sin{\theta})^n = \cos{n\theta}+i\sin{n\theta}$$

The Attempt at a Solution

To find a sum for the series it is a GP with first term,$$a=2^{-1}z$$ common ration,$$r=2^{-1}z$$
then $$S_N = \frac{2^{-1}z(1-(2^{-1}z)^{N})}{1-2^{-1}z}$$

For the second part I was thinking to just replace $$z^n$$ with $$\sin(\frac{1}{10}n\pi)$$ would that work?(NOTE:Also, even though I think I typed the LATEX thing correctly it doesn't display what i actually typed when i previewed the post, so if something looks weird please check if I typed it correctly,such as
\frac{1025\sin(\frac{1}{10}\pi)}{2560-2048\cos(\frac{1}{10}\pi) appears as $$a^3<-9b-3c-3$$

 

[dextercioby]

You're missing a } at the end. EDIT:Not anymore.

Suggestion. $z=e^{i\pi/10}$.

LATER EDIT: Yes, it's already visible in post #1.

 

[rock.freak667]

ah...strange things appear in my browser...$z=e^{i\pi/10}$...so simple...shall try it now

 

[ima_shah2002]

ah...strange things appear in my browser...$z=e^{i\pi/10}$...so simple...shall try it now

Did u get the answer...can u post the solution...

 

[eumyang]

Even if the OP found the answer (almost 4 years ago!), if you need help with the same question, he can't just post the solution. That would be against forum rules. YOU need to post your attempt FIRST, and then maybe we can help.