Many JEE problems involve summing series of inverse tangent terms. The key identity: tan^(-1)(n+1) - tan^(-1)(n) = tan^(-1)(1/(1+n(n+1))). This means tan^(-1)(1/(1+n(n+1))) telescopes when summed. More generally, if a term can be written as tan^(-1)(f(n)) = tan^(-1)(g(n+1)) - tan^(-1)(g(n)) for some function g, the sum telescopes. Common patterns: tan^(-1)(1/(n^2+n+1)), tan^(-1)(2/(n^2)), tan^(-1)((n+1-n)/(1+n(n+1))). To solve: (1) recognize the general term as a difference of two inverse tangents, (2) write out the first few and last few terms to identify cancellation, (3) evaluate the telescoped result. The answer typically involves tan^(-1)(N) - tan^(-1)(1) or similar clean expressions.