BITSAT2019MathematicsInverse Trigonometric FunctionsActual
∑ m = 1 n tan - 1 2 m m 4 + m 2 + 2 is equal to
Options
- Atan - 1 n 2 + n n 2 + n + 2
- Btan - 1 n 2 - n n 2 - n + 2
- Ctan - 1 n 2 + n + 2 n 2 + n
- DNone of these
Correct answer
A. tan - 1 n 2 + n n 2 + n + 2
Step-by-step solution
We have, ∑ m = 1 n tan - 1 2 m m 4 + m 2 + 2 = ∑ m = 1 n tan - 1 m 2 + m + 1 - m 2 - m + 1 1 + m 2 + m + 1 m 2 - m + 1 = ∑ 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 n 2 + n + 1 - 1 1 + n 2 + n + 1 = tan - 1 n 2 + n n 2 + n + 2