Many JEE problems involve summing series of inverse tangent terms. The key identity: tan^(-1)(n+1) - tan^(-1)(n) = tan^(-1)($\frac{1}{1+n(n+1}$)). This means tan^(-1)($\frac{1}{1+n(n+1}$)) 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)($\frac{1}{n^2+n+1}$), tan^(-1)($\frac{2}{n^2}$), tan^(-1)$\frac{(n+1-n}{(1+n(n+1)}$)). 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.