If ∑ r = 1 50 tan - 1 1 2 r 2 = p , then the value of tan p is :

Step-by-step solution

∑ r = 1 50 tan - 1 1 2 r 2 = ∑ r = 1 50 tan - 1 2 4 r 2 = ∑ r = 1 50 tan - 1 2 1 + 4 r 2 - 1 = ∑ r = 1 50 tan - 1 2 1 + 2 r - 1 2 r + 1 = ∑ r = 1 50 tan - 1 ( 2 r + 1 ) - ( 2 r - 1 ) 1 + ( 2 r + 1 ) ( 2 r - 1 ) = ∑ r = 1 50 [ tan - 1 2 r + 1 - tan - 1 2 r - 1 ] So, ∑ r = 1 50 tan - 1 1 2 r 2 = tan - 1 3 - tan - 1 1 + tan - 1 5 - tan - 1 3 + tan - 1 7 - tan - 1 5 . . . . . . . . . . . tan - 1 101 - tan - 1 99 = tan - 1 101 - tan - 1 1 = tan - 1 101 - 1 1 + 101 = tan - 1 100