If tan ^{-1}left(frac{1}{3}right) + tan ^{-1} | vedclass

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(D) The general term of the series is $T_r = 	an ^{-1}\left(\frac{1}{1 + r(r+1)}ight)$.Using the identity $	an ^{-1}(x) - 	an ^{-1}(y) = 	an ^{-

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$12$

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(D) The general term of the series is $T_r = 	an ^{-1}\left(\frac{1}{1 + r(r+1)}ight)$.Using the identity $	an ^{-1}(x) - 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x-y}{1+xy}ight)$,we can write $T_r = 	an ^{-1}(r+1) - 	an ^{-1}(r)$.Summing from $r=1$ to $r=5$:$S = \sum_{r=1}^{5} (	an ^{-1}(r+1) - 	an ^{-1}(r)) = (	an ^{-1} 2 - 	an ^{-1} 1) + (	an ^{-1} 3 - 	an ^{-1} 2) + \dots + (	an ^{-1} 6 - 	an ^{-1} 5)$.This is a telescoping series,so $S = 	an ^{-1} 6 - 	an ^{-1} 1$.$S = 	an ^{-1}\left(\frac{6-1}{1+6 	imes 1}ight) = 	an ^{-1}\left(\frac{5}{7}ight)$.Thus,$p = 5$ and $q = 7$. Since $p$ and $q$ are relatively prime,$p + q = 5 + 7 = 12$.

If $\tan ^{-1}\left(\frac{1}{3}\right) + \tan ^{-1}\left(\frac{1}{7}\right) + \tan ^{-1}\left(\frac{1}{13}\right) + \tan ^{-1}\left(\frac{1}{21}\right) + \tan ^{-1}\left(\frac{1}{31}\right) = \tan ^{-1}\left(\frac{p}{q}\right)$,where $p$ and $q$ are relatively prime numbers,then $p + q$ is equal to-

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