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If \(\mathrm{S}\) is the sum of all the numbers of the form \(\frac{1}{\mathrm{n}}\), where \(\mathrm{n}\) is an integer from \(33\) to \(64\) inclusive, then \(\mathrm{S}\) lies in which of the following intervals?
Let's break down what we're being asked to find. We need to calculate the sum S, which means adding up all the fractions of the form \(\frac{1}{n}\) where n goes from 33 to 64.
So we're looking at: \(\mathrm{S} = \frac{1}{33} + \frac{1}{34} + \frac{1}{35} + \frac{1}{36} + ... + \frac{1}{63} + \frac{1}{64}\)
First, let's figure out how many terms we have. From 33 to 64 inclusive, we count: \(64 - 33 + 1 = 32\) terms total.
Now, rather than trying to add all 32 fractions exactly (which would be very tedious), let's notice that the answer choices are spread out quite a bit. This suggests we can use estimation techniques to find the right interval.
Process Skill: TRANSLATE - Converting the problem language into a clear mathematical sum with a specific number of terms
Here's the key insight: since we have 32 positive fractions to add up, and the answer choices are well-separated intervals, we don't need the exact sum. We can find upper and lower boundaries that will pinpoint which interval contains our answer.
Think of it this way - imagine you're estimating the cost of 32 different items at a store. If you want a rough range, you could multiply 32 by the price of the cheapest item (for a lower bound) and 32 by the price of the most expensive item (for an upper bound).
In our sum, the largest fraction is \(\frac{1}{33}\) (smallest denominator = largest value) and the smallest fraction is \(\frac{1}{64}\) (largest denominator = smallest value).
To get a conservative lower estimate, let's imagine all 32 terms were as small as the smallest term, which is \(\frac{1}{64}\).
Lower bound = \(32 \times \frac{1}{64} = \frac{32}{64} = \frac{1}{2}\)
This means S must be greater than \(\frac{1}{2}\), since our actual sum includes terms larger than \(\frac{1}{64}\).
To get a conservative upper estimate, let's imagine all 32 terms were as large as the largest term, which is \(\frac{1}{33}\).
Upper bound = \(32 \times \frac{1}{33} = \frac{32}{33}\)
Since \(\frac{32}{33}\) is just slightly less than 1 (because \(\frac{33}{33} = 1\)), our upper bound is just under 1.
This means S must be less than \(\frac{32}{33}\), which is less than 1.
Process Skill: INFER - Recognizing that we can sandwich the true sum between clearly calculated bounds
Now we can combine our bounds: \(\frac{1}{2} < \mathrm{S} < \frac{32}{33}\)
Since \(\frac{32}{33}\) is approximately 0.97, we have: \(\frac{1}{2} < \mathrm{S} < 1\) (approximately)
Let's check this against our answer choices:
Our bounds show that \(\frac{1}{2} < \mathrm{S} < 1\), which corresponds exactly to answer choice (D).
This makes intuitive sense: we're adding 32 positive fractions, where each is between \(\frac{1}{64}\) and \(\frac{1}{33}\). The sum should be substantial (more than \(\frac{1}{2}\)) but not huge (less than 1), which is exactly what we found.
Answer: (D) \(\frac{1}{2} < \mathrm{S} < 1\)
1. Misunderstanding the range of terms: Students may incorrectly calculate the number of terms in the sum. When counting integers from 33 to 64 inclusive, they might forget the "inclusive" part and calculate \(64 - 33 = 31\) terms instead of \(64 - 33 + 1 = 32\) terms. This error affects the entire bounding strategy.
2. Attempting exact calculation instead of estimation: Students may try to calculate the exact sum by adding all 32 fractions \(\frac{1}{33} + \frac{1}{34} + ... + \frac{1}{64}\) instead of recognizing that the well-separated answer choices suggest an estimation approach using bounds would be more efficient and sufficient.
3. Confusion about which fraction is larger: Students may incorrectly think that \(\frac{1}{64}\) is larger than \(\frac{1}{33}\) because \(64 > 33\), forgetting that for unit fractions (fractions with numerator 1), a larger denominator means a smaller fraction value.
1. Incorrect bound calculations: When calculating the lower bound, students might compute \(32 \times \frac{1}{64}\) incorrectly, perhaps getting \(\frac{32}{64} = \frac{1}{2}\) but then forgetting that this represents a lower bound (meaning \(\mathrm{S} > \frac{1}{2}\)) rather than thinking \(\mathrm{S} = \frac{1}{2}\).
2. Arithmetic errors in fraction operations: Students may make basic arithmetic mistakes when calculating \(\frac{32}{64} = \frac{1}{2}\) for the lower bound or \(\frac{32}{33}\) for the upper bound, potentially getting confused with fraction simplification or decimal approximations.
1. Boundary interpretation errors: After correctly finding that \(\frac{1}{2} < \mathrm{S} < \frac{32}{33} \approx 1\), students might select choice (C) \(\frac{1}{32} < \mathrm{S} < \frac{1}{2}\) by confusing which side of \(\frac{1}{2}\) their sum falls on, or they might select choice (E) \(1 < \mathrm{S} < 2\) by misinterpreting that the upper bound \(\frac{32}{33}\) is close to but less than 1.
2. Misreading answer choices: Students may correctly determine their bounds but then misread the answer choices, particularly confusing \(\frac{1}{32}\) with \(\frac{1}{2}\) in choices (B) and (C), or not carefully noting the direction of the inequalities in each choice.