Loading...
If \(\mathrm{P}\) equals the product of all 20 numbers of the forum \(\frac{\mathrm{n}}{\mathrm{n} + 1}\), where n is an integer from 1 to 20, inclusive, then \(\mathrm{P} =\)
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:
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
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:
\(\mathrm{P} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times 19 \times 20}{2 \times 3 \times 4 \times 5 \times 6 \times \ldots \times 20 \times 21}\)
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:
Process Skill: VISUALIZE - Seeing the cancellation pattern by organizing the terms systematically
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:
\(\mathrm{P} = \frac{1}{21}\)
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).
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 \(\frac{20}{21}\).
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.
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.
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.
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.