Telescoping product with consecutive odd numerators and denominators Compute (2 − 1/3)(2 − 3/5)(2 − 5/7) … (2 − 997/999).

Introduction / Context:This product is crafted to telescope. Each factor uses consecutive odd integers in the fraction, enabling massive cancellation when expressed with a common pattern.

Given Data / Assumptions:

• General term: 2 − k/(k + 2), where k runs over odd integers 1, 3, 5, …, 997.
• All denominators are nonzero and positive.

Concept / Approach:Rewrite each factor: 2 − k/(k + 2) = (2(k + 2) − k)/(k + 2) = (k + 4)/(k + 2). This turns the entire product into a chain of ratios that cancel successively.

Step-by-Step Solution:For k = 1: (k + 4)/(k + 2) = 5/3.For k = 3: 7/5. For k = 5: 9/7. … For k = 997: 1001/999.Product P = (5/3) × (7/5) × (9/7) × … × (1001/999).Everything cancels except the smallest denominator 3 and the largest numerator 1001.Hence P = 1001/3.

Verification / Alternative check:Write out the first few and last few terms to see direct cancellation of intermediate numerators and denominators.

Why Other Options Are Wrong:1001/999 and 999/1001 reflect partial cancellation; 5/1001 and 3/1001 invert endpoints incorrectly.

Common Pitfalls:Losing the pattern of consecutive odds or not expressing each factor as (k + 4)/(k + 2) before multiplying.

Final Answer:1001/3