\[\displaystyle \sum^{2016}_{n=1}\dfrac{1}{2^{n-1}}\tan \dfrac{\theta}{2^{n-1}}\]

If the value of the above sum is in the form \(\frac{1}{2^{k}}\cot \frac{\theta}{2^{k}}-2\cot m\theta\), where \(k\), \(m \in \mathbb{N}\), then find \(k+m\).

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