If \( \displaystyle \sum_{r=1}^{50} \tan^{-1}\!\left(\frac{1}{2r^{2}}\right) = p \), then the value of \( \tan p \) is:
\( \frac{51}{50} \)
\( \frac{50}{51} \)
\( \frac{101}{102} \)
\( 100 \)