The series of differences between consecutive prime numbers is denoted as Cp1, Cp2, Cp3, ..., Cpn, where Cp1 is the difference between the second and the first prime numbers. If the 23rd prime number is 83, what is the sum of the series Cp1 + Cp2 + ... + Cp22?

Introduction / Context:
This question deals with prime numbers and the differences between consecutive primes. The series Cp1, Cp2, ..., represents the gaps between each pair of consecutive prime numbers. You are given that the 23rd prime is 83 and asked to find the sum of the first 22 such differences. This is a telescoping series idea, where many intermediate terms cancel out, leaving a simple result involving the first and last primes considered.

Given Data / Assumptions:

• Prime numbers in order are p1, p2, p3, ..., p23, where p23 = 83.
• Cp1 = p2 − p1, Cp2 = p3 − p2, ..., Cp22 = p23 − p22.
• We are asked to find Cp1 + Cp2 + ... + Cp22.
• The first prime p1 is 2.

Concept / Approach:
The key concept is that the sum of consecutive differences telescopes. Each Cp term is the difference between two consecutive primes. When we add these differences in order, almost all intermediate primes cancel out, and we are left with the difference between the last and the first prime involved. This type of simplification is called a telescoping sum and is very powerful in series problems.

Step-by-Step Solution:
Step 1: Write the definitions. Cp1 = p2 − p1, Cp2 = p3 − p2, ..., Cp22 = p23 − p22. Step 2: Consider the sum S = Cp1 + Cp2 + ... + Cp22. Step 3: Substitute the definitions: S = (p2 − p1) + (p3 − p2) + ... + (p23 − p22). Step 4: Observe that this is a telescoping series. The +p2 from Cp1 cancels with the −p2 from Cp2, +p3 cancels with −p3, and so on. Step 5: After cancellation, only the first negative term and the last positive term remain: S = p23 − p1. Step 6: The first prime p1 is 2, and the 23rd prime p23 is given as 83. Therefore S = 83 − 2 = 81.

Verification / Alternative check:
You can verify the telescoping idea with a smaller example. Take the first 5 primes: 2, 3, 5, 7, 11. Then Cp1 = 3 − 2 = 1, Cp2 = 5 − 3 = 2, Cp3 = 7 − 5 = 2, Cp4 = 11 − 7 = 4. The sum Cp1 + Cp2 + Cp3 + Cp4 = 1 + 2 + 2 + 4 = 9, which equals 11 − 2, the last prime minus the first. This matches the pattern used in the main problem, giving confidence that S = 83 − 2 = 81 is correct for Cp1 through Cp22.

Why Other Options Are Wrong:
Option 87, 83, 85 and 79: These values might be obtained by guessing or by misinterpreting the series definition (for example, adding one more term or using a wrong starting or ending prime). However, once the telescoping structure is correctly applied, the only consistent result is 81.

Common Pitfalls:
A common mistake is to misunderstand n and think that Cp23 should also be included in the sum, or to forget that Cp1 is defined as the difference between the second and the first prime. Another error is trying to list all primes and explicitly compute each difference, which is time consuming and unnecessary here. Recognising the telescoping pattern S = last prime − first prime is the key to solving such problems quickly.

Final Answer:
The sum of the first 22 differences between consecutive prime numbers up to the 23rd prime is 81.