If _r=1^50^-1(12r^2)=p, then p is:

Question

If $\displaystyle\sum_{r=1}^{50}	an^{-1}\left(\frac1{2r^2}ight)=p$, then $	an p$ is:

Accepted Answer

$$\tan^{-1}\left(\frac1{2r^2}\right) = \tan^{-1}\left(\frac{2}{1+4r^2-1}\right) = \tan^{-1}(2r+1)-\tan^{-1}(2r-1)$$

Telescoping: $\sum_{r=1}^{50} = \tan^{-1}(101)-\tan^{-1}(1) = \tan^{-1}\left(\frac{101-1}{1+101}\right) = \tan^{-1}\frac{50}{51}$

$$\tan p = \frac{50}{51}$$

Solutions · 1