Geome(trig)tric progression

Applying logarithm and differentiating the identity $\displaystyle \prod_{r=1}^{n}\cos \frac{\theta}{2^r} = \frac{\sin \theta}{2^n\sin \frac{\theta}{2^n}}$ we get , $\displaystyle \sum_{r=1}^{n} \frac{1}{2^r}\tan \frac{\theta}{2^r} = \frac{1}{2^n}\cot \frac{\theta}{2^n}-\cot \theta$

So, $\displaystyle \sum_{n=1}^{2016} \frac{1}{2^{n-1}} \tan \frac{\theta}{2^{n-1}} = \frac{1}{2^{2015}}\tan \frac{\theta}{2^{2015}} -\cot\theta+\tan\theta = \frac{1}{2^{2015}}\tan \frac{\theta}{2^{2015}} - 2\cot 2\theta$ which makes answer $\boxed{2+2015=2017}$

Since $\cot\theta - 2\cot2\theta \; = \; \tan\theta$ this series telescopes, with $S \; = \; \sum_{n=1}^{2016}\left[\frac{1}{2^{n-1}}\cot\left(\frac{\theta}{2^{n-1}}\right) - \frac{1}{2^{n-2}}\cot\left(\frac{\theta}{2^{n-2}}\right) \right] \; =\;\frac{1}{2^{2015}}\cot\left(\frac{\theta}{2^{2015}}\right) - 2\cot2\theta$ making the answer $2015 + 2 = \boxed{2017}$ .

Let the sum be $S$

Then $(1/2^{2016} )(cot(\theta/2^{2016}) -S= 2cot(2\theta)$ ,

By a property that $-tan(\theta)+ cot(\theta)=2cot(2\theta)$

This property gets applyied succesively as soon as last term of $-S$ is paired with $(1/2^{2016} )(cot(\theta/2^{2016})$ . Thus answer follows.