The objective is to determine the value of \( x \) satisfying the equation:
\(\tan^{-1}\left(\frac{1}{1+1\cdot2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}(x)\)
Observe that each term in the series can be expressed in the form:
\(\tan^{-1}\left(\frac{1}{1+k(k+1)}\right) = \tan^{-1}\left(\frac{1}{k^2+k+1}\right)\)
Utilize the arctangent addition formula:
\(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)\) for \( ab < 1 \).
Rewrite each term using the identity relating tangent differences:
\(\tan(\theta) = \frac{\tan A+\tan B}{1-\tan A \tan B}\)
Reformulate the problem using the additive property of arctangent:
\(\tan^{-1}((k + 1) - k) = \tan^{-1}(\text{term})\)
This leads to a telescoping series of the form:
\(\tan^{-1}(1) - \tan^{-1}\left(\frac{1}{n+1}\right)\)
After cancellation of intermediate terms, the sum simplifies to:
\(\tan^{-1}\left(\frac{n}{n+2}\right)\) by rewriting the general term as:
\(\frac{1}{1+k(k+1)} = \frac{(k+1)-k}{1 + k(k+1)}\)
The final simplified form of the series is:
\(\tan^{-1}(x) = \tan^{-1}\left(\frac{n}{n+2}\right)\)
Therefore, \( x = \frac{n}{n+2} \), consistent with the problem statement.