Let S = 1/(1 × 3 × 5) + 1/(1 × 4) + 1/(3 × 5 × 7) + 1/(4 × 7) + 1/(5 × 7 × 9) + 1/(7 × 10) + ... up to 20 terms. What is the value of S?

Introduction / Context:
This is a challenging series simplification problem from higher level aptitude exams. The series S consists of alternating terms with three factors in the denominator and two factors in the denominator. The denominators follow a clear pattern, and the sum up to 20 terms must be evaluated exactly as a rational number.

Given Data / Assumptions:
We are given:

• S = 1/(1 × 3 × 5) + 1/(1 × 4) + 1/(3 × 5 × 7) + 1/(4 × 7) + 1/(5 × 7 × 9) + 1/(7 × 10) + ...
• The pattern continues in the same way up to 20 terms.
• The terms alternate between a fraction with three consecutive odd numbers and a fraction with two numbers that form an arithmetic progression with common difference 3.

Concept / Approach:
The key idea is to identify a general term for the sequence and then express each term using partial fraction decomposition so that the overall series telescopes. Telescoping means that many intermediate terms cancel out when we sum the series, leaving only a few boundary terms to evaluate. This technique is common in advanced series problems.

Step-by-Step Solution:
Observe the triple denominator terms: (1,3,5), (3,5,7), (5,7,9), ... These form consecutive odd numbers. For the n-th triple term, the denominator can be written as (2n - 1)(2n + 1)(2n + 3). Similarly, the double denominator terms are (1,4), (4,7), (7,10), ... which follow the pattern (3n - 2, 3n + 1). Hence S can be written as the sum from n = 1 to 10 of two terms for each n: 1/[(2n - 1)(2n + 1)(2n + 3)] + 1/[(3n - 2)(3n + 1)]. Now use partial fractions for the triple term: 1/[(2n - 1)(2n + 1)(2n + 3)] can be expressed as a difference of two fractions with denominators (2n - 1)(2n + 1) and (2n + 1)(2n + 3). Similarly, 1/[(3n - 2)(3n + 1)] can be decomposed into a difference involving 1/(3n - 2) and 1/(3n + 1). When we write out several terms explicitly, most intermediate fractions cancel in a telescoping manner, leaving only a few starting and ending contributions. Carrying this algebra through up to n = 10 gives a final exact fraction S = 6070/14973.

Verification / Alternative check:
We can verify numerically by computing all 20 terms using a calculator or a computer algebra system. Evaluating the exact rational sum confirms that S equals 6070/14973, which also matches the option given.

Why Other Options Are Wrong:
6179/15275, 7191/15174 and 5183/16423 are close looking rational numbers but result from incorrect partial fraction decomposition or an incorrect count of terms. Because the cancellation pattern is subtle, losing or adding one extra term at either end of the telescoping series easily leads to one of these incorrect distractors. The option 4501/12025 is an additional plausible fraction but does not match the carefully computed exact sum.

Common Pitfalls:
Students may misread the pattern and assume pure arithmetic or geometric progression instead of the given alternating structure. Another common mistake is to stop telescoping too early or mishandle the indices for the first and last terms, which leads to a slightly wrong rational result. Writing down the first few terms clearly and checking the cancellation process step by step is essential.

Final Answer:
The value of S up to 20 terms is 6070/14973.