Question

$\sum_{m = 1}^{n}\tan^{- 1}\frac{2m}{m^{4} + m^{2} + 2}$=

• A. tan^–1 (n² + n + 1)
• B. tan^–1 (n² – n + 1)
• C. tan^–1$\frac{n^{2} + n}{n^{2} + n + 2}$
• D. None of these

Answer 1

Answer: (C)

tan^–1$\frac{n^{2} + n}{n^{2} + n + 2}$

Answer 2

tan^–1 $\frac{2m}{m^{4} + m^{2} + 2}$ = tan^–1

$\frac{2m}{1 + (m^{2} + m + 1)(m^{2} - m + 1)}$

̃ tan^–1 (m² + m + 1) – tan^–1 (m² – m + 1)

So that $\left( \frac{2m}{m^{4} + m^{2} + 2} \right)$

= (tan^–1 3 – tan^–1 1) + (tan^–1 7 – tan^–1 3) + …. + tan^–1(n² + n +

1) – tan^–1 (n² – n + 1)

̃ tan^–1 (n² + n + 1) – tan^–1 (1)

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

= (tan^–1 3 – tan^–1) + (tan^–1 7 – tan^–1 3) + …..+

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

= tan^–1 (n² + n + 1) – tan^–1 (n² – n + 1)

= tan^–1 $\frac{n^{2} + n}{n^{2} + n + 2}$.