Question

$\tan^{-1} \left( \frac{1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \cdot 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n \cdot (n+1)} \right) =$

• A. $\tan^{-1} \left( \frac{n}{n+2} \right)$
• B. $\tan^{-1} \left( \frac{n+1}{n} \right)$
• C. $\tan^{-1} \left( \frac{n}{n+1} \right)$
• D. $\tan^{-1} \left( \frac{n+2}{n} \right)$

Hint:

For series summation involving $\tan^{-1}$, always try to express the argument in the form $\frac{x-y}{1+xy}$. This allows you to split the term into $\tan^{-1}x - \tan^{-1}y$, which almost always leads to a telescoping series where most terms neatly cancel out.

Answer 1

The Correct Option is A