Concept:The series telescopes using the identity tan−1a−tan−1b=tan−1(1+aba−b) after rewriting each denominator as (x+r)(x+r−1)+1.Explanation:The r-th denominator is x2+(2r−1)x+r(r−1)+1, which equals (x+r)(x+r−1)+1.So the r-th term becomes tan−1((x+r)(x+r−1)+11).Put a=x+r and b=x+r−1 in the identity tan−1a−tan−1b=tan−1(1+aba−b).Here, a−b=1 and 1+ab=1+(x+r)(x+r−1).Thus, tan−1(x+r)−tan−1(x+r−1)=tan−1((x+r)(x+r−1)+11).This exactly matches the r-th term of the given series.Therefore, y=[tan−1(x+1)−tan−1x]+[tan−1(x+2)−tan−1(x+1)]+⋯+[tan−1(x+n)−tan−1(x+n−1)].All middle terms cancel, so y=tan−1(x+n)−tan−1x.Differentiating, y′=1+(x+n)21−1+x21.At x=0:y′(0)=1+n21−1=n2+11−(n2+1)=−n2+1n2.Answer:y′(0)=−n2+1n2, which is Option B.