For any positive integer n, define fn:(0,∞)→R as fn(x) = n∑j=1narctan((1+(x+j)(x+j))-1) for all x∈(0,∞). (Here, the inverse trigonometric function arctan x assumes values in (-π/2, π/2)). Then, which of the following statement(s) is (are) TRUE?

The correct options are A and D.
5∑j=15tan2(fj(0))=55
For any fixed positive integer n, limx→∞sec2(fn(x))=1
fn(x) = n∑j=1narctan((x+j)-1 - (x+j)-1) / (1+(x+j)(x+j))
fn(x) = n∑j=1n[arctan(x+j) - arctan(x+j)]
fn(x) = arctan(x+n) - arctan(x)
∴tan(fn(x)) = tan[arctan(x+n) - arctan(x)]
tan(fn(x)) = (x+n) - x / 1+x(x+n)
tan(fn(x)) = n / 1+x2+nx
tan(fn(0)) = n
tan2(fn(0)) = n2
5∑n=05tan(fn(0)) = 5∑n=05n2 = 55
∴sec2(fn(x)) = 1+tan2(fn(x))
sec2(fn(x)) = 1+(n/(1+x2+nx))2
∴limx→∞sec2(fn(x)) = 1