If n is a positive integer, then \((-2^{\mathrm{n}})^{-2}+(2^{-\mathrm{n}})^2 =\)

1. Translate the problem requirements: We need to simplify the expression \((-2^\mathrm{n})^{-2}+(2^{-\mathrm{n}})^2\) where n is a positive integer, recognizing that this involves negative exponents and parentheses that affect the order of operations.
2. Simplify each term independently: Apply exponent rules to each of the two terms separately, being careful about the placement of negative signs and parentheses.
3. Combine the simplified terms: Add the two simplified expressions together and look for common factors or further simplification opportunities.
4. Match with answer choices: Compare the final simplified form with the given options to identify the correct answer.

Execution of Strategic Approach

1. Translate the problem requirements

We need to simplify the expression \((-2^\mathrm{n})^{-2}+(2^{-\mathrm{n}})^2\) where n is a positive integer. Let's break down what this means in plain English:

• We have two separate terms that we need to add together
• The first term is \((-2^\mathrm{n})^{-2}\), which means we take \((-2^\mathrm{n})\) and raise it to the power of -2
• The second term is \((2^{-\mathrm{n}})^2\), which means we take \((2^{-\mathrm{n}})\) and square it
• We need to be very careful about the order of operations and where the negative signs are placed

Process Skill: TRANSLATE - Converting the complex notation into understandable components

2. Simplify each term independently

Let's work on each term separately using basic exponent rules.

First term: \((-2^\mathrm{n})^{-2}\)
In plain English: We're taking \((-2^\mathrm{n})\) and raising it to the power of -2.

• First, let's understand what \(-2^\mathrm{n}\) means: this is \(-(2^\mathrm{n})\), so if \(\mathrm{n}=3\), we'd have \(-(2^3) = -8\)
• Now we raise this to the power of -2: \([-(2^\mathrm{n})]^{-2}\)
• Using the rule that \((\mathrm{a}^\mathrm{m})^\mathrm{p} = \mathrm{a}^{\mathrm{mp}}\), we get: \([-(2^\mathrm{n})]^{-2} = (2^\mathrm{n})^{-2} \times (-1)^{-2}\)
• Since \((-1)^{-2} = \frac{1}{(-1)^2} = \frac{1}{1} = 1\)
• So we have: \((2^\mathrm{n})^{-2} = 2^{\mathrm{n} \times (-2)} = 2^{-2\mathrm{n}}\)

Second term: \((2^{-\mathrm{n}})^2\)
In plain English: We're taking \((2^{-\mathrm{n}})\) and squaring it.

• Using the rule \((\mathrm{a}^\mathrm{m})^\mathrm{p} = \mathrm{a}^{\mathrm{mp}}\): \((2^{-\mathrm{n}})^2 = 2^{(-\mathrm{n}) \times 2} = 2^{-2\mathrm{n}}\)

Process Skill: SIMPLIFY - Breaking down complex expressions using basic exponent rules

3. Combine the simplified terms

Now we add our two simplified terms together:

• First term simplified to: \(2^{-2\mathrm{n}}\)
• Second term simplified to: \(2^{-2\mathrm{n}}\)
• Adding them: \(2^{-2\mathrm{n}} + 2^{-2\mathrm{n}}\)

In plain English: We have the same term twice, so we're adding it to itself.

• This gives us: \(2 \times 2^{-2\mathrm{n}}\)
• Using the rule that \(\mathrm{a} \times \mathrm{b}^\mathrm{c} = \mathrm{b}^\mathrm{c} \times \mathrm{a}\), we can write this as: \(2^1 \times 2^{-2\mathrm{n}}\)
• Using the rule that \(\mathrm{a}^\mathrm{m} \times \mathrm{a}^\mathrm{n} = \mathrm{a}^{\mathrm{m}+\mathrm{n}}\): \(2^1 \times 2^{-2\mathrm{n}} = 2^{1 + (-2\mathrm{n})} = 2^{1 - 2\mathrm{n}} = 2^{-2\mathrm{n} + 1}\)

Process Skill: MANIPULATE - Combining like terms and applying exponent addition rules

4. Final Answer

Our final simplified expression is \(2^{-2\mathrm{n}+1}\).

Looking at the answer choices:

1. \(0\) ❌
2. \(2^{-2\mathrm{n}}\) ❌
3. \(2^{2\mathrm{n}}\) ❌
4. \(2^{-2\mathrm{n}+1}\) ✓
5. \(2^{2\mathrm{n}+1}\) ❌

The answer is D. \(2^{-2\mathrm{n}+1}\)

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the order of operations with negative signs
Students often confuse the meaning of \(-2^\mathrm{n}\) versus \((-2)^\mathrm{n}\). In this problem, \(-2^\mathrm{n}\) means \(-(2^\mathrm{n})\), not \((-2)^\mathrm{n}\). This confusion can lead to incorrect interpretation of the first term \((-2^\mathrm{n})^{-2}\), causing students to treat it as \([(-2)^\mathrm{n}]^{-2}\) instead of \([-(2^\mathrm{n})]^{-2}\).

2. Not recognizing that both terms need to be simplified separately before combining
Some students might try to combine the terms algebraically before simplifying each exponential expression individually. This approach leads to confusion and incorrect manipulation of the exponents and negative signs.

3. Overlooking that the final result will involve adding like terms
Students may not anticipate that both terms will simplify to the same base expression (\(2^{-2\mathrm{n}}\)), missing the opportunity to recognize this as a 'combine like terms' situation early in their approach.

Errors while executing the approach

1. Incorrect handling of negative exponents
When simplifying \((-2^\mathrm{n})^{-2}\), students often make errors with the negative exponent rules. They might incorrectly calculate \((-1)^{-2}\) or forget that \(\mathrm{a}^{-\mathrm{m}} = \frac{1}{\mathrm{a}^\mathrm{m}}\), leading to sign errors in their intermediate steps.

2. Arithmetic mistakes when applying the power rule \((\mathrm{a}^\mathrm{m})^\mathrm{n} = \mathrm{a}^{\mathrm{mn}}\)
Students frequently make calculation errors when multiplying exponents, such as calculating \((-\mathrm{n}) \times 2\) incorrectly, or mixing up signs when computing \(\mathrm{n} \times (-2) = -2\mathrm{n}\) for the first term.

3. Error in combining the final terms using exponent addition rules
When adding \(2^{-2\mathrm{n}} + 2^{-2\mathrm{n}} = 2 \times 2^{-2\mathrm{n}}\), students often struggle with rewriting 2 as \(2^1\) and then applying the rule \(2^1 \times 2^{-2\mathrm{n}} = 2^{1 + (-2\mathrm{n})} = 2^{-2\mathrm{n} + 1}\). They might incorrectly add exponents or make sign errors.

Errors while selecting the answer

1. Confusing the final exponent form
Students might arrive at the correct numerical relationship but write the exponent in the wrong form. For example, they might choose \(2^{1-2\mathrm{n}}\) instead of recognizing it's equivalent to \(2^{-2\mathrm{n}+1}\), or vice versa, leading them to think their answer doesn't match any of the given choices.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for n

Let's choose \(\mathrm{n} = 2\) (a small positive integer that will make calculations manageable)

Step 2: Substitute n = 2 into the original expression

\((-2^\mathrm{n})^{-2} + (2^{-\mathrm{n}})^2\) becomes:

\((-2^2)^{-2} + (2^{-2})^2\)

Step 3: Evaluate each term with n = 2

First term: \((-2^2)^{-2} = (-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}\)

Second term: \((2^{-2})^2 = (\frac{1}{4})^2 = \frac{1}{16}\)

Step 4: Add the terms

\(\frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}\)

Step 5: Express 1/8 in terms of powers of 2

\(\frac{1}{8} = 2^{-3}\)

Step 6: Check which answer choice gives \(2^{-3}\) when n = 2

1. \(0 \neq 2^{-3}\)
2. \(2^{-2\mathrm{n}} = 2^{-4} = \frac{1}{16} \neq 2^{-3}\)
3. \(2^{2\mathrm{n}} = 2^4 = 16 \neq 2^{-3}\)
4. \(2^{-2\mathrm{n}+1} = 2^{-4+1} = 2^{-3}\) ✓
5. \(2^{2\mathrm{n}+1} = 2^{4+1} = 2^5 = 32 \neq 2^{-3}\)

Step 7: Verify with another value (n = 1)

With \(\mathrm{n} = 1\): \((-2^1)^{-2} + (2^{-1})^2 = (-2)^{-2} + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 2^{-1}\)

Checking answer D: \(2^{-2(1)+1} = 2^{-1}\) ✓

Answer: D