$ = \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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$(A)$ $f(x)$ is monotonically increasing on $[1, \infty)$
$(B)$ $f(x)$ is monotonically decreasing on $(0,1)$
$(C)$ $f(x)+f\left(\frac{1}{x}\right)=0$, for all $x \in(0, \infty)$
$(D)$ $f\left(2^x\right)$ is an odd function of $x$ on $R$