JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The general term of the given series is T_p = ^ -1 ( 2^ p-1 1 + 2^ 2p-1 ). This can be rewritten by expressing the numerator and denominator in terms of 2^p and 2^ p-1 : T_p = ^ -1 ( 2^p - 2^ p-1 1 + 2^p 2^ p-1 ) Using the inverse trigonometric identity ^ -1 ( x - y 1 + xy ) = ^ -1 x - ^ -1 y, we get: T_p = ^ -1 (2^p) - ^ -1 (2^ p-1 ) Summing T_p from p = 1 to 11 gives a telescoping series: _ p=1 ^ 11 T_p = ( ^ -1 (2^1) - ^ -1 (2^0) ) + ( ^ -1 (2^2) - ^ -1 (2^1) ) + + ( ^ -1 (2^ 11 ) - ^ -1 (2^ 10 ) ) Canceling the intermediate terms, we obtain: _ p=1 ^ 11 T_p = ^ -1 (2^ 11 ) - ^ -1 (2^0) Since 2^0 = 1 and ^ -1 (1) = 4 , the sum becomes: _ p=1 ^ 11 T_p = ^ -1 (2^ 11 ) - 4 Substituting this back into the expression for : = 4 + ^ -1 (2^ 11 ) - 4 = ^ -1 (2^ 11 ) Taking the tangent of both sides, we get: = 2^ 11 = 2048 Answer: 2048