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

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(C) We are given the sum $p = \sum_{r=1}^{50} 	an ^{-1} \frac{1}{2 r^2}$.Multiply the numerator and denominator inside the $	an^{-1}$ by $2$ to get

Vedclass · Accepted Answer

$\frac{50}{51}$

Vedclass · Answer

(C) We are given the sum $p = \sum_{r=1}^{50} 	an ^{-1} \frac{1}{2 r^2}$.Multiply the numerator and denominator inside the $	an^{-1}$ by $2$ to get $	an ^{-1} \frac{2}{4 r^2}$.We can rewrite the expression as $	an ^{-1} \left[ \frac{(2r+1) - (2r-1)}{1 + (2r+1)(2r-1)} ight]$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$,this becomes $	an^{-1}(2r+1) - 	an^{-1}(2r-1)$.Now,the sum is a telescoping series:$p = \sum_{r=1}^{50} [	an^{-1}(2r+1) - 	an^{-1}(2r-1)]$$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}$:$p = 	an^{-1} \left( \frac{101 - 1}{1 + 101 	imes 1} ight) = 	an^{-1} \left( \frac{100}{102} ight)$.Therefore,$	an p = \frac{100}{102} = \frac{50}{51}$.

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

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