Question

The sum of series $\left( \frac { 2 k } { 2 + k ^ { 2 } + k ^ { 4 } } \right)$

• A. tan^–1 (n² + n + 1)
• B. $\frac { \pi } { 4 }$– tan^–1 (n² – n + 1)
• C. $\frac { \pi } { 4 }$ + cot^–1 (n² + n + 1)
• D. None of these

Answer 1

Answer: (C)

$\frac { \pi } { 4 }$ + cot^–1 (n² + n + 1)

Answer 2

= $\sum _ { \mathrm { k } = 1 } ^ { \mathrm { n } } \left[ \tan ^ { - 1 } \left( \mathrm { k } ^ { 2 } + \mathrm { k } + 1 \right) - \tan ^ { - 1 } \left( \mathrm { k } ^ { 2 } - \mathrm { k } + 1 \right) \right]$= (tan^–1 3 – tan^–11) + (tan^–1 7 – tan^–1 3) + (tan^–1 13 – tan^–1 7) +…..

+ [tan^–1 (n² + n + 1) – tan^–1(n² – n + 1)]

= tan^–1 (n² + n + 1) – tan^–1(1)

= tan^–1(n² + n + 1) – p/4

= $\frac { \pi } { 2 }$– cot^–1 (n² + n + 1) – $\frac { \pi } { 4 }$ = $\frac { \pi } { 4 }$ – cot^–1(n² + n + 1)