Evaluate the telescoping surd series (1 - √2) + (√2 - √3) + (√3 - √4) + … + (√15 - √16), simplify it completely, and then select the correct final value from the options.

Introduction / Context:
This problem involves a telescoping series of surds. In such expressions, many intermediate terms cancel out when the series is written in full. Recognising the telescoping pattern greatly simplifies the work and allows you to compute the sum quickly without individually evaluating each square root with a calculator.

Given Data / Assumptions:

• The expression is (1 - √2) + (√2 - √3) + (√3 - √4) + … + (√15 - √16).
• The pattern continues in steps of 1 under the square root from 2 up to 16.
• All roots are principal non negative square roots.
• We sum all the terms exactly, without decimal approximation.

Concept / Approach:
A telescoping sum is one in which intermediate terms cancel when the series is written out explicitly. Here each term has the form √n - √(n + 1). When you add successive terms, the positive √(n + 1) from one term cancels with the negative √(n + 1) from the next term. Only the very first and the very last square root terms survive in the final sum.

Step-by-Step Solution:
Write out the first few terms explicitly: (1 - √2) + (√2 - √3) + (√3 - √4) + (√4 - √5) + … + (√15 - √16). Observe the cancellations: -√2 cancels with +√2, -√3 cancels with +√3, and so on, all the way up to √15. After cancellation, only the very first 1 and the very last term -√16 remain. Because √16 = 4, the sum simplifies to 1 - 4 = -3.
Verification / Alternative check:
You can check by grouping terms in partial sums, for example up to (√5 - √6) or (√7 - √8), and you will see the same pattern of cancellation. Each time, intermediate square roots vanish, leaving only the first and last terms in the chain. This behaviour confirms that the telescoping interpretation is correct.

Why Other Options Are Wrong:
Option b ( 0 ) would require the initial 1 to be cancelled as well, which never happens. Option c ( 1 ) ignores the contribution of the last term -√16. Option d ( 4 ) and option e ( -4 ) both treat the series as if only the largest square root mattered, which is not accurate because the initial 1 also remains in the final result.

Common Pitfalls:
A common mistake is to stop the series too early or misidentify the last term under the square root, leading to an incorrect final expression. Another pitfall is to approximate each surd numerically and then add them, which is unnecessary and prone to rounding errors. Recognising the telescoping nature makes the problem much simpler.

Final Answer:
The value of the telescoping surd series is -3.