How do i find sum $\sum_{r=1}^{n}\sin(3^{r-1}\theta)\sec(3^r\theta) $ How do i find sum 
$$\sum_{r=1}^{n}\sin(3^{r-1}\theta)\sec(3^r\theta) $$
Needs hint to begin
thanks

Note that this can be done through significant amounts of trigonemetric manipulation on $\frac{\sin x}{\cos 3x}$. So first,
$$\frac{\sin x}{\cos 3x}= \frac{1}{2} \cdot \frac{2 \cdot \sin x \cdot \cos x}{\cos x \cdot \cos 3x}= \frac{1}{2} \cdot  \frac{\sin 2x}{\cos x \cdot \cos 3x}$$
Now, this becomes
$$ \frac{1}{2} \cdot \frac{ \sin(3x - x) }{ \cos x \cdot \cos3x }= \frac{1}{2} \cdot \frac{ \sin 3x \cdot \cos x - \cos 3x \cdot \sin x}{ \cos x \cdot \cos3x}$$
Which is $$ \frac{1}{2} \cdot \frac {\sin 3x \cdot \cos x}{\cos x \cdot \cos 3x} - \frac{1}{2} \cdot \frac{\cos 3x \cdot \sin x}{\cos x \cdot \cos 3x}= \frac{1}{2} \cdot \left (\frac{\sin 3x}{\cos 3x} - \frac{\sin x}{\cos x} \right )=\frac{\tan 3x-\tan x}{2}$$
So we have $$\frac{\sin x}{\cos 3x}=\frac{\tan 3x-\tan x}{2} \tag{1}$$
So $$\sum_{r=1}^{n}\sin(3^{r-1}\theta)\sec(3^r\theta) =\sum_{r=1}^{n}\frac{\sin(3^{r-1}\theta)}{\cos(3^r\theta)}=\frac{1}{2} \left( \sum_{r=1}^{n}\tan 3^r \theta-\tan 3^{r-1} \theta \right)$$From $(1)$. Note that this is a telescoping series. You can simplify this nicely. 
