If sum_{r=1}^{50} tan ^{-1} frac{1}{2 r^{2}}= | vedclass

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(D) We are given the sum $p = \sum_{r=1}^{50} 	an ^{-1} \frac{1}{2 r^{2}}$.First,multiply the numerator and denominator by $2$ to use the formula $

Vedclass · Accepted Answer

$\frac{50}{51}$

Vedclass · Answer

(D) We are given the sum $p = \sum_{r=1}^{50} 	an ^{-1} \frac{1}{2 r^{2}}$.First,multiply the numerator and denominator by $2$ to use the formula $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \frac{x-y}{1+xy}$.$p = \sum_{r=1}^{50} 	an ^{-1} \frac{2}{4 r^{2}} = \sum_{r=1}^{50} 	an ^{-1} \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$.This is a telescoping series of the form $\sum_{r=1}^{n} (	an ^{-1}(a_{r+1}) - 	an ^{-1}(a_r))$.$p = (	an ^{-1} 3 - 	an ^{-1} 1) + (	an ^{-1} 5 - 	an ^{-1} 3) + \dots + (	an ^{-1} 101 - 	an ^{-1} 99)$.All intermediate terms cancel out,leaving $p = 	an ^{-1} 101 - 	an ^{-1} 1$.Using the formula $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \frac{x-y}{1+xy}$,we get $p = 	an ^{-1} \frac{101-1}{1+(101)(1)} = 	an ^{-1} \frac{100}{102} = 	an ^{-1} \frac{50}{51}$.Therefore,$	an p = \frac{50}{51}$.

If $\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^{2}}=p$,then the value of $\tan p$ is :

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