What I'm getting is $$\frac{( \sin (N+1)x - 2^N \sin x)}{(2^N(\sin x - 2))}$$
How do I simplify to the form they have given , please help.
I hope it's clear because I don't know Ajax still thats why I avoided putting the picture, please if any of you find pictures aren't appropriate , just give me some days to learn Ajax.
After this ending point I just simplified what i got.
$$\sum _{n=1}^N \frac{\sin(\theta n)}{2^n}$$
First, let's rewrite it as an exponential ->
$$\sum _{n=1}^N i 2^{-n-1} e^{-i \theta n}-i 2^{-n-1} e^{i \theta n}$$
Sum->
$$ = \frac{i2^{-N-1}e^{-i\theta(N+1)}\left(2e^{i\theta}-e^{2i\theta}-2^{N+1}e^{i \theta(N+1)}+e^{2i\theta(N+1)}+2^{N+1}e^{2i\theta+i\theta(N+1)}-2e^{i\theta+2i\theta (N+1)}\right)}{\left(-2+e^{i\theta}\right)\left(-1+2e^{i\theta}\right)}$$
Reduce ->
$$ = \frac{i 2^{-N-1} e^{-i \theta N} \left(-e^{i \theta }-2^{N+1} e^{i \theta N}-2 e^{2 i \theta (N+1)}+2^{N+1} e^{i \theta (N+2)}+e^{i (\theta +2 \theta N)}+2\right)}{-\left(5 e^{i \theta }\right)+2 e^{2 i \theta }+2}$$
Expand, factor, and reduce ->
$$ = -\frac{2^{-N} \left(\sin (\theta N)-2 (\sin (\theta (N+1)))+2^{N+1} (\theta \sin )\right)}{4 (\theta \cos )-5} $$
Re-arrange ->
$$ = \frac{\sin (\theta N)-2 (\sin (\theta (N+1)))+2^{N+1} (\theta \sin )}{2^N (4 (\theta \cos )-5)}$$
Swap the signs ->
$$ = \frac{\sin (\theta N)+2 (\sin (\theta (N+1)))-2^{N+1} (\theta \sin )}{2^N (5-4 (\theta \cos ))}$$