If $\sum_{r=1}^{50} an ^{-1} \frac{1}{2 r^2}=p$ then $ an p$ is

Question

If $\sum_{r=1}^{50} 	an ^{-1} \frac{1}{2 r^2}=p$ then $	an p$ is

Aicharya · Accepted Answer

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

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

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