$$\sum\limits_{m=1}^{n}{{{\tan }^{-1}}}\left( \frac{2m}{{{m}^{4}}+{{m}^{2}}+2} \right)$$is equal to

Question

TuteeHUB · Accepted Answer

$${{\tan }^{-1}}\left( \frac{{{n}^{2}}-n}{{{n}^{2}}-n+2} \right)$$

TuteeHUB · Answer

$${{\tan }^{-1}}\left( \frac{{{n}^{2}}+n}{{{n}^{2}}+n+2} \right)$$

TuteeHUB · Answer

$${{\tan }^{-1}}\left( \frac{{{n}^{2}}+n+2}{{{n}^{2}}+n} \right)$$

TuteeHUB · Answer

None of these

1.	\[\sum\limits_{m=1}^{n}{{{\tan }^{-1}}}\left( \frac{2m}{{{m}^{4}}+{{m}^{2}}+2} \right)\]is equal to
A.	\[{{\tan }^{-1}}\left( \frac{{{n}^{2}}+n}{{{n}^{2}}+n+2} \right)\]
B.	\[{{\tan }^{-1}}\left( \frac{{{n}^{2}}-n}{{{n}^{2}}-n+2} \right)\]
C.	\[{{\tan }^{-1}}\left( \frac{{{n}^{2}}+n+2}{{{n}^{2}}+n} \right)\]
D.	None of these
Answer» B. \[{{\tan }^{-1}}\left( \frac{{{n}^{2}}-n}{{{n}^{2}}-n+2} \right)\]

\[\sum\limits_{m=1}^{n}{{{\tan }^{-1}}}\left( \frac{2m}{{{m}^{4}}+{{m}^{2}}+2} \right)\]is equal to