sumlimits_{m = 1}^n {{{tan }^{ - 1}}} left( { | vedclass

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(A) We have $\sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} ight)} $$= \sum\limits_{m = 1}^n {{{	an }^{ - 1}}\lef

Vedclass · Accepted Answer

${	an ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} ight)$

Vedclass · Answer

(A) We have $\sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} ight)} $$= \sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{2m}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} ight)} $$= \sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{({m^2} + m + 1) - ({m^2} - m + 1)}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} ight)} $$= \sum\limits_{m = 1}^n {[{{	an }^{ - 1}}({m^2} + m + 1) - {{	an }^{ - 1}}({m^2} - m + 1)]} $$= ({	an ^{ - 1}}3 - {	an ^{ - 1}}1) + ({	an ^{ - 1}}7 - {	an ^{ - 1}}3) + ({	an ^{ - 1}}13 - {	an ^{ - 1}}7) + \dots + [{	an ^{ - 1}}({n^2} + n + 1) - {	an ^{ - 1}}({n^2} - n + 1)]$$= {	an ^{ - 1}}({n^2} + n + 1) - {	an ^{ - 1}}1$$= {	an ^{ - 1}}\left( {\frac{{({n^2} + n + 1) - 1}}{{1 + ({n^2} + n + 1)(1)}}} ight) = {	an ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} ight)$.

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

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