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If \(\mathrm{a} = 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64}\) and \(\mathrm{b} = 1 + \frac{1}{4}\mathrm{a}\), then what is the value of \(\mathrm{a} - \mathrm{b}\)?
Let's understand what we're being asked to find. We have two values:
- '\(\mathrm{a}\)' is the sum of four specific fractions: \(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64}\)
- '\(\mathrm{b}\)' is defined as \(1 + \frac{1}{4}\) times the value of '\(\mathrm{a}\)'
Our goal is to find the difference \(\mathrm{a} - \mathrm{b}\). This means we need to calculate both '\(\mathrm{a}\)' and '\(\mathrm{b}\)' first, then subtract '\(\mathrm{b}\)' from '\(\mathrm{a}\)'.
Process Skill: TRANSLATE - Converting the problem statement into clear mathematical requirements
Now let's find the exact value of \(\mathrm{a}\) by adding up all the fractions. To add fractions, we need a common denominator.
Looking at our fractions: \(1\), \(\frac{1}{4}\), \(\frac{1}{16}\), and \(\frac{1}{64}\), we can see that:
- \(1 = \frac{64}{64}\)
- \(\frac{1}{4} = \frac{16}{64}\)
- \(\frac{1}{16} = \frac{4}{64}\)
- \(\frac{1}{64} = \frac{1}{64}\)
So: \(\mathrm{a} = \frac{64}{64} + \frac{16}{64} + \frac{4}{64} + \frac{1}{64}\)
Adding the numerators: \(\mathrm{a} = \frac{64 + 16 + 4 + 1}{64} = \frac{85}{64}\)
Now we can find '\(\mathrm{b}\)' using the formula \(\mathrm{b} = 1 + \frac{1}{4}\mathrm{a}\).
We know \(\mathrm{a} = \frac{85}{64}\), so:
\(\mathrm{b} = 1 + \frac{1}{4} \cdot \frac{85}{64}\)
\(\mathrm{b} = 1 + \frac{85}{256}\)
To add these, we need a common denominator:
\(\mathrm{b} = \frac{256}{256} + \frac{85}{256} = \frac{256 + 85}{256} = \frac{341}{256}\)
Now we can calculate \(\mathrm{a} - \mathrm{b}\):
\(\mathrm{a} - \mathrm{b} = \frac{85}{64} - \frac{341}{256}\)
To subtract these fractions, we need a common denominator. Since \(256 = 4 \times 64\), we can write:
\(\frac{85}{64} = \frac{340}{256}\)
Therefore:
\(\mathrm{a} - \mathrm{b} = \frac{340}{256} - \frac{341}{256} = \frac{340 - 341}{256} = -\frac{1}{256}\)
The value of \(\mathrm{a} - \mathrm{b} = -\frac{1}{256}\), which matches answer choice B.
Let's verify: We found \(\mathrm{a} = \frac{85}{64}\) and \(\mathrm{b} = \frac{341}{256}\). Converting \(\mathrm{a}\) to the same denominator: \(\mathrm{a} = \frac{340}{256}\). So \(\mathrm{a} - \mathrm{b} = \frac{340}{256} - \frac{341}{256} = -\frac{1}{256}\). ✓
Students may misread the expression \(\mathrm{b} = 1 + \frac{1}{4}\mathrm{a}\) as \(\mathrm{b} = \left(1 + \frac{1}{4}\right)\mathrm{a}\), which would mean \(\mathrm{b} = \frac{5}{4}\mathrm{a}\). This changes the entire calculation and leads to a completely different answer. The parentheses and order of operations are crucial here.
Some students might recognize that \(\mathrm{a}\) looks like it could be part of a geometric series (with first term 1 and ratio \(\frac{1}{4}\)), but may try to apply the infinite series formula when we only have 4 specific terms. This can lead to using \(\mathrm{a} = \frac{1}{1-\frac{1}{4}} = \frac{4}{3}\) instead of calculating the actual sum.
When converting fractions to have denominator 64, students often make mistakes such as writing \(\frac{1}{4} = \frac{4}{64}\) instead of \(\frac{16}{64}\), or \(\frac{1}{16} = \frac{1}{64}\) instead of \(\frac{4}{64}\). These errors in fraction conversion lead to an incorrect value of \(\mathrm{a}\).
Students may incorrectly calculate \(\frac{1}{4} \times \frac{85}{64}\) as \(\frac{85}{16}\) instead of \(\frac{85}{256}\), forgetting that they need to multiply both the numerator by numerator and denominator by denominator: \(\frac{1 \times 85}{4 \times 64} = \frac{85}{256}\).
When computing \(\mathrm{a} - \mathrm{b}\), students need to express both fractions with the same denominator. They might incorrectly convert \(\frac{85}{64}\) to \(\frac{85}{256}\) instead of \(\frac{340}{256}\), or make errors when performing \(\frac{340}{256} - \frac{341}{256}\).
No likely faltering points - the final calculation directly gives \(-\frac{1}{256}\), which matches exactly with answer choice B, leaving little room for selection errors.