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
Correct Answer: A
Solution :
We have \[\sum\limits_{m=1}^{n}{{{\tan }^{-1}}\left( \frac{2m}{{{m}^{4}}+{{m}^{2}}+2} \right)}\] \[=\sum\limits_{m=1}^{n}{{{\tan }^{-1}}\left( \frac{2m}{1+({{m}^{2}}+m+1)({{m}^{2}}-m+1)} \right)}\] \[=\sum\limits_{m=1}^{n}{{{\tan }^{-1}}\left( \frac{({{m}^{2}}+m+1)-({{m}^{2}}-m+1)}{1+({{m}^{2}}+m+1)({{m}^{2}}-m+1)} \right)}\] =\[\sum\limits_{m=1}^{n}{[{{\tan }^{-1}}({{m}^{2}}+m+1)-{{\tan }^{-1}}({{m}^{2}}-m+1)]}\] \[=({{\tan }^{-1}}3-{{\tan }^{-1}}1)+({{\tan }^{-1}}7-{{\tan }^{-1}}3)+\] \[({{\tan }^{-1}}13-{{\tan }^{-1}}7)+......+[{{\tan }^{-1}}({{n}^{2}}+n+1)\]\[-{{\tan }^{-1}}({{n}^{2}}-n+1)]\] = \[{{\tan }^{-1}}({{n}^{2}}+n+1)-{{\tan }^{-1}}1\]= \[{{\tan }^{-1}}\left( \frac{{{n}^{2}}+n}{2+{{n}^{2}}+n} \right)\].
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