The given series is:

$S = \sum_{k=2}^{512} \frac{1}{\sqrt{2k} + \sqrt{2k+2}}$

Step 1: Rationalizing the Denominators

We multiply the numerator and denominator by the conjugate of the denominator:

$\frac{1}{\sqrt{2k} + \sqrt{2k+2}} \times \frac{\sqrt{2k+2} - \sqrt{2k}}{\sqrt{2k+2} - \sqrt{2k}}$

This simplifies the denominator:

$(\sqrt{2k+2} + \sqrt{2k})(\sqrt{2k+2} - \sqrt{2k}) = (2k+2) - (2k) = 2$

Thus, each term simplifies to:

$\frac{\sqrt{2k+2} - \sqrt{2k}}{2}$

Step 2: Recognizing the Telescoping Sum

Rewriting the sum:

$S = \sum_{k=2}^{512} \frac{\sqrt{2k+2} - \sqrt{2k}}{2}$

This is a telescoping series where intermediate terms cancel out, leaving only the first and last terms:

$S = \frac{1}{2} (\sqrt{1024} - \sqrt{4})$

Step 3: Final Calculation

Since $\sqrt{1024} = 32$ and $\sqrt{4} = 2$, we get:

$S = \frac{1}{2} (32 - 2) = \frac{30}{2} = 15$

Conclusion:

$S = 15$

Would you like me to explain any step in more detail?

Related Questions:

1. How does the method of telescoping sums help simplify series?
2. What happens if the series starts at a different value of $k$?
3. How does rationalizing denominators simplify fraction calculations?
4. What are other examples of telescoping sums in mathematics?
5. Can this method be applied to a similar sum involving cube roots?

Tip:

Always check for patterns in series—many complicated sums can simplify significantly with algebraic manipulation!