Question

The sum up to \( n \) terms of the series \( \frac{1}{\sqrt{1} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{11}} + \cdots \) is:

• A. \( \frac{1}{5} \left[ \sqrt{5n+1} \right] \)
• B. \( \frac{1}{5} \left[ \sqrt{5n+1} + 1 \right] \)
• C. \( \frac{1}{5} \left[ \sqrt{5n+1} - 1 \right] \)
• D. \( \frac{1}{6} \left[ \sqrt{6n+1} \right] \)
• E. \( \frac{1}{7} \left[ \sqrt{7n+1} - 1 \right] \)

Hint:

For telescoping series, write the general term in a form where consecutive terms cancel out, simplifying the sum significantly.

Answer 1

The Correct Option is C

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), \( \frac{1}{5} \left[ \sqrt{5n+1} - 1 \right] \).