If P equals the product of all 20 numbers of the forum n/(n + 1), where n is an integer...

1. Translate the problem requirements: We need to find the product P of all 20 fractions of the form \(\frac{\mathrm{n}}{\mathrm{n+1}}\) where n goes from 1 to 20. This means \(\mathrm{P} = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{20}{21}\).
2. Write out the product structure: Express P as a single fraction by multiplying all numerators together and all denominators together to see the overall pattern.
3. Identify the telescoping cancellation pattern: Observe how consecutive terms share common factors that will cancel out when we arrange the numerators and denominators systematically.
4. Apply the cancellation to find the simplified result: Determine what remains after all possible cancellations occur and verify this matches one of the answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We need to find the product P of all fractions where each fraction has the form \(\frac{\mathrm{n}}{\mathrm{n+1}}\), and n goes from 1 to 20.

This means we're multiplying:

• First fraction: \(\frac{1}{2}\) (when n=1)
• Second fraction: \(\frac{2}{3}\) (when n=2)
• Third fraction: \(\frac{3}{4}\) (when n=3)
• And so on...
• Last fraction: \(\frac{20}{21}\) (when n=20)

So our product P looks like:

\(\mathrm{P} = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \ldots \times \frac{19}{20} \times \frac{20}{21}\)

Process Skill: TRANSLATE - Converting the mathematical notation into concrete fractions we can work with

2. Write out the product structure

When we multiply fractions, we multiply all the numerators together and all the denominators together. Let's organize this to see the pattern more clearly.

Numerators: \(1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times 19 \times 20\)

Denominators: \(2 \times 3 \times 4 \times 5 \times 6 \times \ldots \times 20 \times 21\)

So we can write:

\(\mathrm{P} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times 19 \times 20}{2 \times 3 \times 4 \times 5 \times 6 \times \ldots \times 20 \times 21}\)

3. Identify the telescoping cancellation pattern

Now here's where the magic happens! Let's look at what appears in both the numerator and denominator:

In the numerator we have: 1, 2, 3, 4, 5, ..., 19, 20

In the denominator we have: 2, 3, 4, 5, 6, ..., 20, 21

Notice that almost everything cancels out:

• The 2 in the numerator cancels with the 2 in the denominator
• The 3 in the numerator cancels with the 3 in the denominator
• The 4 in the numerator cancels with the 4 in the denominator
• And this continues all the way up to...
• The 20 in the numerator cancels with the 20 in the denominator

Process Skill: VISUALIZE - Seeing the cancellation pattern by organizing the terms systematically

4. Apply the cancellation to find the simplified result

After all the cancellations, what's left?

From the numerator: Only the 1 remains (since 1 doesn't appear in the denominator)

From the denominator: Only the 21 remains (since 21 doesn't appear in the numerator)

Therefore: \(\mathrm{P} = \frac{1}{21}\)

Let's verify this makes sense by checking our logic:

• Every number from 2 to 20 appears exactly once in both numerator and denominator, so they all cancel
• Only 1 (from numerator) and 21 (from denominator) don't have matching partners
• Result: \(\frac{1}{21}\)

5. Final Answer

\(\mathrm{P} = \frac{1}{21}\)

This matches answer choice C: \(\frac{1}{21}\)

The telescoping product works because each fraction shares a common factor with the next fraction, creating a chain of cancellations that leaves only the very first numerator (1) and the very last denominator (21).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the product notation

Students may struggle to understand what "product of all 20 numbers of the form \(\frac{\mathrm{n}}{\mathrm{n+1}}\)" means. They might think they need to find a general formula for \(\frac{\mathrm{n}}{\mathrm{n+1}}\) rather than multiplying 20 specific fractions together. This leads to confusion about whether to work with the general form or write out all individual fractions from \(\frac{1}{2}\) to \(\frac{20}{21}\).

2. Missing the telescoping pattern opportunity

Even when students correctly identify that they need to multiply fractions, they may not recognize this as a telescoping product. Instead, they might attempt to calculate the product by finding a common denominator or by multiplying out all numerators and denominators without looking for cancellation patterns, making the problem unnecessarily complex.

Errors while executing the approach

1. Incomplete cancellation identification

When applying the telescoping cancellation, students often miss some of the canceling terms or fail to systematically match numerators with denominators. For example, they might cancel 2 through 10 but forget that numbers 11 through 20 also appear in both numerator and denominator and should cancel out completely.

2. Arithmetic errors in organizing terms

Students may make errors when writing out what remains in the numerator versus denominator after cancellation. They might incorrectly think that multiple terms remain uncanceled, or conversely, they might think everything cancels out completely, leading to incorrect final fractions.

Errors while selecting the answer

1. Confusing the final simplified result

After correctly identifying that only 1 remains in the numerator and 21 in the denominator, students might second-guess themselves and select a more "complex-looking" answer like \(\frac{1}{20 \times 21}\) thinking their solution was "too simple." The elegance of the telescoping result (just \(\frac{1}{21}\)) may seem too straightforward compared to the other answer choices.