If $\sum_{r=1}^{50} \tan^{-1}\frac{1}{2r^{2 = p}}$ then $\tan p$ is
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Recognize patterns that allow for telescoping sums, especially with inverse trigonometric functions. The key is to manipulate the argument of the $\tan^{-1}$ function into the form $\frac{A-B}{1+AB}$ so it can be expressed as $\tan^{-1}A - \tan^{-1}B$.