_ m = 1 ^n ^ - 1 ( 2m m^4 + m^2 + 2 ) is equal to - Vidyadip

MCQ

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

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

Correct option: A.

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

a
(a) 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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