To solve the given series:
\(\tan^{-1} \left( \frac{1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \cdot 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n \cdot (n+1)} \right)\)
Let's start by rewriting the general term:
Each term in the series is of the form \(\tan^{-1} \left( \frac{1}{1 + k(k+1)} \right)\) where \( k \) ranges from 1 to \( n \).
The denominator simplifies to \(1 + k(k+1) = k^2 + k + 1\), but let's focus on possible simplification:
\(\frac{1}{1 + k(k+1)} = \frac{1}{k^2 + k + 1}\)
Rewrite the fraction:
\(\frac{1}{1 + k(k+1)} = \frac{1}{(k+1)^2 - k^2} = \frac{1}{(k+1)^2 - k^2}\)
Recognize this as a form that matches the identity for the inverse tangent:
\(\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x+y}{1-xy} \right)\)
Apply the identity:
The identity you can use is:
\(\tan^{-1} \left( \frac{1}{k} \right) - \tan^{-1} \left( \frac{1}{k+2} \right) = \tan^{-1}\left( \frac{(k+2) - k}{1 + \frac{1}{k(k+2)}} \right)\)
For our sum:
Each term can be rewritten as \(\tan^{-1} \left( \frac{1}{k} \right) - \tan^{-1} \left( \frac{1}{k+2} \right)\), which will telescope when the series is summed over.
The telescoping nature simplifies:
The terms cancel out sequentially:
\(\tan^{-1} \left(1\right) - \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{3} \right) - \tan^{-1} \left( \frac{1}{5} \right) + \dots + \tan^{-1} \left( \frac{1}{n+1} \right) - \tan^{-1} \left( \frac{1}{n+3} \right)\)
Thus, all intermediate terms cancel out:
Only the first and last parts remain: \(\tan^{-1} \left(1\right) - \tan^{-1} \left( \frac{1}{n+3} \right)\)
Simplify the result:
\(\tan^{-1}\left( \frac{(n+3)-1}{1+\frac{1}{n+3}} \right) = \tan^{-1}\left( \frac{n+2}{1} \right) = \tan^{-1}(n+2)\)
Thus, the final expression becomes:
\(\tan^{-1} \left( \frac{n}{n+2} \right)\)
The correct answer is:
\(\tan^{-1} \left( \frac{n}{n+2} \right)\)