Evaluate the telescoping series by partial fractions: 1/(1*4) + 1/(4*7) + 1/(7*10) + 1/(10*13) + 1/(13*16) = ?

Introduction / Context:This series uses a fixed step in the denominator, suggesting partial fractions of the form 1/(n(n + 3)) = (1/3)(1/n − 1/(n + 3)). Such decompositions create telescoping sums in which most intermediate terms cancel cleanly.

Given Data / Assumptions:

• Terms: n = 1, 4, 7, 10, 13 with step 3.
• Series: Σ 1/[n(n + 3)] over these five n.
• Exact arithmetic; reduce fractions to simplest form.

Concept / Approach:Decompose each term: 1/(n(n + 3)) = (1/3)(1/n − 1/(n + 3)). Summing these creates successive cancellations, leaving only the first 1/n and the very last −1/(n + 3).

Step-by-Step Solution:

Verification / Alternative check:Compute numerically and compare: 0.25 + 0.035714... + 0.014285... + 0.007692... + 0.004807... ≈ 0.3125 = 5/16.

Why Other Options Are Wrong:3/16 and 7/16 result from off-by-one cancellations; 11/16 is far too large; 1/3 ignores the finite truncation at 1/16.

Common Pitfalls:Forgetting the 1/3 factor in the decomposition or misaligning the start/end indices, which ruins cancellation.

Final Answer:5/16