If \(\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}=p\) then \(\tan p\) is

\(\begin{aligned} & \sum_{\mathrm{r}=1}^{50} \tan ^{-1} \frac{1}{2 \mathrm{r}^2}=\mathrm{p} \\ & \Rightarrow \sum_{\mathrm{r}=1}^{50} \tan ^{-1}\left(\frac{2}{4 \mathrm{r}^2}\right)=\mathrm{p} \\ & \Rightarrow \sum_{\mathrm{r}=1}^{50} \tan ^{-1}\left[\frac{(2 \mathrm{r}+1)-(2 \mathrm{r}-1)}{1+(2 \mathrm{r}+1)(2 \mathrm{r}-1)}\right]=\mathrm{p} \\ & \Rightarrow \sum_{\mathrm{r}=1}^{50}\left[\tan ^{-1}(2 \mathrm{r}+1)-\tan ^{-1}(2 \mathrm{r}-1)\right]=\mathrm{p} \\ & \Rightarrow \tan ^{-1}(101)-\tan ^{-1}(1)=\mathrm{p} \\ & \Rightarrow \tan ^{-1}\left(\frac{101-1}{1+101}\right)=\mathrm{p} \\ & \Rightarrow \frac{100}{102}=\tan \mathrm{p} \\ & \Rightarrow \tan \mathrm{p}=\frac{50}{51}\end{aligned}\)