$ = \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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Given below are two statements:
Statement I: $f(-x)$ is the inverse of the matrix $f(x)$.
Statement II: $f(x) f(y)=f(x+y)$.
In the light of the above statements, choose the correct answer from the options given below