Kerala KEAM 5 June 2024 Solved Paper

We are given the series: $$S_n = \frac{1}{\sqrt{1} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{11}} + \cdots$$We need to find the sum of the first $n$ terms of this series.The general term of the series can be written as: $$T_k = \frac{1}{\sqrt{5(k-1)+1} + \sqrt{5k+1}}$$where $k$ is the term number.To simplify this, multiply and divide the expression by the conjugate of the denominator: $$T_k = \frac{1}{\sqrt{5(k-1)+1} + \sqrt{5k+1}} \cdot \frac{\sqrt{5k+1} - \sqrt{5(k-1)+1}}{\sqrt{5k+1} - \sqrt{5(k-1)+1}}$$This simplifies to: $$T_k = \frac{\sqrt{5k+1} - \sqrt{5(k-1)+1}}{(\sqrt{5k+1})^2 - (\sqrt{5(k-1)+1})^2}$$ $$T_k = \frac{\sqrt{5k+1} - \sqrt{5(k-1)+1}}{(5k+1) - (5(k-1)+1)}$$ $$T_k = \frac{\sqrt{5k+1} - \sqrt{5(k-1)+1}}{5}$$So, the sum of the first $n$ terms is: $$S_n = \sum_{k=1}^{n} \frac{1}{5} \left( \sqrt{5k+1} - \sqrt{5(k-1)+1} \right)$$This is a telescoping series, where most of the terms cancel out, leaving: $$S_n = \frac{1}{5} \left( \sqrt{5n+1} - 1 \right)$$Thus, the correct answer is option (C), $\frac{1}{5} \left[ \sqrt{5n+1} - 1 \right]$. Quick Tip: For telescoping series, write the general term in a form where consecutive terms cancel out, simplifying the sum significantly.