Concept:This is a telescoping sum using the identity tan−1x−tan−1y=tan−1(1+xyx−y).Explanation:Consider the general term of the series.For k=1,2,…,n, the k-th term is tan−1(1+k(k+1)1).Set x=k+1 and y=k.Then tan−1(k+1)−tan−1k=tan−1(1+k(k+1)(k+1)−k).This becomes tan−1(1+k(k+1)1).Hence each term can be written as tan−1(k+1)−tan−1k.Substituting k=1,2,…,n gives a telescoping chain:(tan−12−tan−11)+(tan−13−tan−12)+⋯+(tan−1(n+1)−tan−1n).All intermediate terms cancel, leaving tan−1(n+1)−tan−11.Apply the same identity again:tan−1(n+1)−tan−11=tan−1(1+(n+1)⋅1(n+1)−1)=tan−1(n+2n).Answer:tan−1(n+2n), which is option A.