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Simplify $\sum\limits_{k = 0}^n {{2^k}\tan ({2^k}x)}$ which $k \in \{ 0,1,...,n + 1\} ,{2^k}x \notin \{ 0,\frac{\pi }{2}\}$

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P.S.: What does segma mean? I'm sure you didn't mean this. But beware of inuendos. ;P – Raskolnikov Mar 2 '11 at 14:45
up vote 11 down vote accepted

Hint: $\ln(\cos(x))'=\tan(x)$ and Recursion.

Alternative route: Use the identity $\cot(x) - 2\cot(2x)=\tan(x)$ and telescoping.

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+1: The first solution is very nice! – Aryabhata Mar 2 '11 at 15:32
thanks a bunch! ^_^ – fidele007 Mar 4 '11 at 16:23

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