Evaluate the telescoping sum: 1/(1·2·3) + 1/(2·3·4) + 1/(3·4·5) + 1/(4·5·6). Express your result as a simplified fraction.

Introduction / Context:
Series with triple products in denominators often telescope when rewritten with partial fractions. Recognizing the pattern 1/(n(n+1)(n+2)) allows rapid summation across consecutive n values. This problem tests that technique on four consecutive terms from n = 1 to n = 4.

Given Data / Assumptions:

• Sum S = 1/(1·2·3) + 1/(2·3·4) + 1/(3·4·5) + 1/(4·5·6).
• All terms are positive rational numbers.
• We seek a single reduced fraction for S.

Concept / Approach:
Use the identity 1/(n(n+1)(n+2)) = 1/2 * [1/(n(n+1)) − 1/((n+1)(n+2))]. Applying this to each term causes consecutive terms to cancel (telescope), leaving only the first and the last boundary terms to evaluate.

Step-by-Step Solution:
Rewrite each term: 1/(n(n+1)(n+2)) = (1/2)[1/(n(n+1)) − 1/((n+1)(n+2))].Sum from n=1 to 4: S = (1/2)[1/(1·2) − 1/(2·3) + 1/(2·3) − 1/(3·4) + 1/(3·4) − 1/(4·5) + 1/(4·5) − 1/(5·6)].Telescoping cancels interior pairs, leaving S = (1/2)[1/(1·2) − 1/(5·6)].Compute: (1/2)[1/2 − 1/30] = (1/2)[(15 − 1)/30] = (1/2)(14/30) = 7/30.

Verification / Alternative check:
Approximate numerically: 1/6 + 1/24 + 1/60 + 1/120 ≈ 0.1667 + 0.0417 + 0.0167 + 0.0083 ≈ 0.2334, and 7/30 ≈ 0.2333 — consistent within rounding.

Why Other Options Are Wrong:

• 11/30 / 13/30 / 17/30 / 2/15: These result from partial cancellations done incorrectly or arithmetic slips in the final subtraction.

Common Pitfalls:
Forgetting the 1/2 factor; misaligning telescoping boundaries; arithmetic errors converting to a common denominator.

Final Answer:
7 / 30