\(\tan^{-1}\left(\dfrac{1}{1 + 1 \cdot 2}\right) + \tan^{-1}\left(\dfrac{1}{1 + 2 \cdot 3}\right) + \ldots + \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right) = \)

The general term of the given series is \(T_k = \tan^{-1}\left(\dfrac{1}{1 + k(k+1)}\right)\).

This can be written as \(T_k = \tan^{-1}\left(\dfrac{(k+1) - k}{1 + (k+1)k}\right)\).

Using the identity \(\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y\), we get:

\(T_k = \tan^{-1}(k+1) - \tan^{-1}k\)

Writing the terms for \(k = 1, 2, \ldots, n\):

\(T_1 = \tan^{-1}2 - \tan^{-1}1\)

\(T_2 = \tan^{-1}3 - \tan^{-1}2\)

\(T_n = \tan^{-1}(n+1) - \tan^{-1}n\)

Adding all the terms, the intermediate terms cancel out:

\(S = \sum_{k=1}^{n} T_k = \tan^{-1}(n+1) - \tan^{-1}1\)

Applying the inverse tangent difference formula again:

\(S = \tan^{-1}\left(\dfrac{(n+1) - 1}{1 + (n+1)(1)}\right)\)

\(S = \tan^{-1}\left(\dfrac{n}{n+2}\right)\)

Answer: \(\tan^{-1}\left(\dfrac{n}{n+2}\right)\)