Which of the following is equivalent to {4^(2n) - 1}{2^(3n) + 1}?

1. Translate the problem requirements: We need to simplify the expression \(\frac{4^{2n - 1}}{2^{3n + 1}}\) and express it in the form \(2^{\text{some power}}\) to match one of the answer choices.
2. Convert to common base: Since 4 = 2², we'll rewrite the numerator using base 2 to have the same base in both numerator and denominator.
3. Apply exponent rules for division: Once we have the same base, we'll use the rule \(\frac{a^m}{a^n} = a^{m-n}\) to combine the exponents.
4. Simplify the resulting exponent: We'll combine like terms in the exponent to get our final answer in the form \(2^{\text{simplified power}}\).

Execution of Strategic Approach

1. Translate the problem requirements

We need to simplify the expression \(\frac{4^{2n - 1}}{2^{3n + 1}}\) and express it in the form \(2^{\text{some power}}\) to match one of the answer choices.

Let's think about what we have here: we're looking at a fraction where the top has 4 raised to some power, and the bottom has 2 raised to some power. Since all our answer choices are powers of 2, we need to get everything in terms of base 2.

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical goal

2. Convert to common base

Since 4 = 2², we'll rewrite the numerator using base 2 to have the same base in both numerator and denominator.

Let's start with the numerator: \(4^{2n-1}\)
Since 4 = 2², we can substitute this in:
\(4^{2n-1} = (2^2)^{2n-1}\)

Now, when we have a power raised to another power, we multiply the exponents together. So:
\((2^2)^{2n-1} = 2^{2 \cdot (2n-1)} = 2^{4n-2}\)

Our expression now becomes:
\(\frac{4^{2n - 1}}{2^{3n + 1}} = \frac{2^{4n-2}}{2^{3n + 1}}\)

3. Apply exponent rules for division

Once we have the same base, we'll use the rule \(\frac{a^m}{a^n} = a^{m-n}\) to combine the exponents.

When we divide powers with the same base, we subtract the exponents. Think of it this way: if you have \(\frac{2^5}{2^3}\), you're dividing \(2 \times 2 \times 2 \times 2 \times 2\) by \(2 \times 2 \times 2\), which leaves you with \(2 \times 2 = 2^2 = 2^{5-3}\).

So for our expression:
\(\frac{2^{4n-2}}{2^{3n + 1}} = 2^{(4n-2) - (3n+1)}\)

Process Skill: MANIPULATE - Applying fundamental exponent rules systematically

4. Simplify the resulting exponent

We'll combine like terms in the exponent to get our final answer in the form \(2^{\text{simplified power}}\).

Let's work out the exponent step by step:
\((4n-2) - (3n+1)\)

First, let's distribute the negative sign:
\(= 4n - 2 - 3n - 1\)

Now, let's group the like terms together:
\(= (4n - 3n) + (-2 - 1)\)
\(= n - 3\)

Therefore: \(\frac{4^{2n - 1}}{2^{3n + 1}} = 2^{n-3}\)

Final Answer

Our simplified expression is \(2^{n-3}\), which matches answer choice A exactly.

Let's verify this makes sense: we started with a more complex expression involving different bases and powers, converted everything to the same base (base 2), and used fundamental exponent rules to simplify. The result \(2^{n-3}\) is in the same form as all our answer choices, confirming our approach was correct.

Answer: A. \(2^{n-3}\)

Common Faltering Points

Errors while devising the approach

1. Not recognizing the need for a common base: Students may attempt to work with the expression \(\frac{4^{2n-1}}{2^{3n+1}}\) directly without converting to a common base. They might try to simplify 4 and 2 separately or look for patterns that don't exist, leading them down an inefficient or incorrect path.

2. Confusion about which base to choose: While the solution converts 4 to base 2, some students might try to convert 2 to base 4 instead (writing 2 as \(4^{1/2}\)), which makes the arithmetic much more complex and error-prone.

Errors while executing the approach

1. Incorrect application of power rules: When converting \(4^{2n-1}\) to \((2^2)^{2n-1}\), students often make the mistake of adding exponents instead of multiplying them, getting \(2^{2+(2n-1)} = 2^{2n+1}\) instead of the correct \(2^{2(2n-1)} = 2^{4n-2}\).

2. Sign errors when subtracting exponents: When applying \(\frac{a^m}{a^n} = a^{m-n}\), students frequently forget to distribute the negative sign properly. They might write \((4n-2) - (3n+1)\) as \(4n-2-3n+1\) instead of the correct \(4n-2-3n-1\), leading to \(n-1\) instead of \(n-3\).

3. Arithmetic errors in combining like terms: Even when the setup is correct, students may make simple arithmetic mistakes when simplifying \((4n-2)-(3n+1)\), such as incorrectly calculating \(-2-1 = -1\) instead of \(-3\), or \(4n-3n = 2n\) instead of \(n\).

Errors while selecting the answer

1. Misreading answer choices: Students might correctly arrive at \(2^{n-3}\) but then select choice C (\(2^{-n+2}\)) or choice E (\(2^{-n-2}\)) due to careless reading, especially since these involve similar-looking negative exponents and variable combinations.