If a sum of money, P dollars, is invested at the interest rate of i% per year, compounded annually, and...

1. Translate the problem requirements: We need to find how many times greater the investment value is after \(3\mathrm{n}\) years compared to after \(2\mathrm{n}\) years, given annual compound interest of \(\mathrm{i}\%\)
2. Set up the compound interest formula: Apply the standard compound interest formula to express the investment value at both time periods
3. Create the ratio relationship: Set up the fraction comparing the \(3\mathrm{n}\)-year value to the \(2\mathrm{n}\)-year value to find the multiplicative factor
4. Simplify using exponential properties: Use the rule that when dividing powers with the same base, we subtract exponents to arrive at the final answer

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We have some initial amount of money \(\mathrm{P}\) dollars that we're investing. This money earns interest at a rate of \(\mathrm{i}\%\) per year, and importantly, this interest gets compounded annually - meaning each year we earn interest not just on our original investment, but also on all the interest we've earned in previous years.

The question wants to know: if we compare how much money we have after \(3\mathrm{n}\) years versus how much we have after \(2\mathrm{n}\) years, what's the ratio? In other words, how many times bigger is our investment after \(3\mathrm{n}\) years compared to after \(2\mathrm{n}\) years?

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical question about ratios of investment values

2. Set up the compound interest formula

When money is compounded annually, there's a standard pattern for how investments grow. Let's think about this step by step:

After 1 year: Our \(\mathrm{P}\) dollars becomes \(\mathrm{P} \times (1 + \mathrm{i}/100)\)
After 2 years: We take last year's amount and multiply by \((1 + \mathrm{i}/100)\) again
After 3 years: We continue this pattern...

The general pattern is that after \(\mathrm{t}\) years, our investment is worth: \(\mathrm{P} \times (1 + \mathrm{i}/100)^\mathrm{t}\)

So specifically:
• Value after \(2\mathrm{n}\) years = \(\mathrm{P} \times (1 + \mathrm{i}/100)^{(2\mathrm{n})}\)
• Value after \(3\mathrm{n}\) years = \(\mathrm{P} \times (1 + \mathrm{i}/100)^{(3\mathrm{n})}\)

3. Create the ratio relationship

Now we need to find how many times greater the \(3\mathrm{n}\)-year value is compared to the \(2\mathrm{n}\)-year value. This means we want to calculate:

Ratio = (Value after \(3\mathrm{n}\) years) ÷ (Value after \(2\mathrm{n}\) years)

Let's substitute what we found:

Ratio = \([\mathrm{P} \times (1 + \mathrm{i}/100)^{(3\mathrm{n})}] \div [\mathrm{P} \times (1 + \mathrm{i}/100)^{(2\mathrm{n})}]\)

Notice that the \(\mathrm{P}\) dollars appears in both the numerator and denominator, so it will cancel out:

Ratio = \((1 + \mathrm{i}/100)^{(3\mathrm{n})} \div (1 + \mathrm{i}/100)^{(2\mathrm{n})}\)

4. Simplify using exponential properties

Here's where we use a key mathematical property: when we divide powers that have the same base, we subtract the exponents.

The rule is: \(\mathrm{a}^\mathrm{m} \div \mathrm{a}^\mathrm{n} = \mathrm{a}^{(\mathrm{m}-\mathrm{n})}\)

Applying this to our ratio:
Ratio = \((1 + \mathrm{i}/100)^{(3\mathrm{n})} \div (1 + \mathrm{i}/100)^{(2\mathrm{n})}\)
Ratio = \((1 + \mathrm{i}/100)^{(3\mathrm{n} - 2\mathrm{n})}\)
Ratio = \((1 + \mathrm{i}/100)^\mathrm{n}\)

Process Skill: SIMPLIFY - Using exponent rules to reduce the expression to its simplest form

4. Final Answer

The investment after \(3\mathrm{n}\) years will be \((1 + \mathrm{i}/100)^\mathrm{n}\) times as great as the value after \(2\mathrm{n}\) years.

Looking at our answer choices, this matches choice C: \((1 + \mathrm{i}/100)^\mathrm{n}\)

We can verify this makes sense: if \(\mathrm{n} = 1\) and \(\mathrm{i} = 10\%\), then after 2 years we'd have some amount, and after 3 years we'd have \((1.10)^1 = 1.10\) times as much, which means 10% more - exactly one additional year of growth, which is logical.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what the question is asking for: Students often think they need to find the actual values after \(2\mathrm{n}\) and \(3\mathrm{n}\) years, rather than understanding that the question asks for the ratio (how many times greater). This leads them to set up calculations for absolute values instead of relative comparison.

2. Confusion about the compound interest setup: Students may forget that interest compounds annually and try to use simple interest formulas, or they may incorrectly think they need to account for the principal amount \(\mathrm{P}\) differently in their setup.

3. Overlooking the algebraic nature of n: Since \(\mathrm{n}\) is described as 'a positive integer' but no specific value is given, students might try to plug in specific numbers instead of recognizing this needs to be solved algebraically with \(\mathrm{n}\) as a variable.

Errors while executing the approach

1. Failing to cancel out the principal amount P: When setting up the ratio \([\mathrm{P} \times (1 + \mathrm{i}/100)^{(3\mathrm{n})}] \div [\mathrm{P} \times (1 + \mathrm{i}/100)^{(2\mathrm{n})}]\), students often forget that \(\mathrm{P}\) appears in both numerator and denominator and should cancel out, leading them to include \(\mathrm{P}\) in their final answer.

2. Incorrect application of exponent rules: When simplifying \((1 + \mathrm{i}/100)^{(3\mathrm{n})} \div (1 + \mathrm{i}/100)^{(2\mathrm{n})}\), students might add the exponents instead of subtracting them, getting \((1 + \mathrm{i}/100)^{(5\mathrm{n})}\) instead of the correct \((1 + \mathrm{i}/100)^\mathrm{n}\).

3. Arithmetic errors with exponent subtraction: Students may make simple calculation mistakes when computing \(3\mathrm{n} - 2\mathrm{n}\), especially if they're working quickly under time pressure.

Errors while selecting the answer

1. Selecting answers that include the principal P: After correctly calculating \((1 + \mathrm{i}/100)^\mathrm{n}\), students might second-guess themselves and choose answer choices D or E that include \(\mathrm{P}\), thinking they 'must have missed something' since their answer seems 'too simple.'

2. Confusing fractional exponents: Students might be tempted by answer choice B with the exponent \(3/2\), thinking it relates to the ratio of time periods (\(3\mathrm{n}\) to \(2\mathrm{n}\)), without recognizing that this doesn't follow from the proper mathematical derivation.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient smart numbers
Let's select specific values that make calculations simple:
• \(\mathrm{P} = \$1000\) (initial investment)
• \(\mathrm{i} = 20\%\) (interest rate)
• \(\mathrm{n} = 2\) (positive integer)

This gives us:
• Time period 1: \(2\mathrm{n} = 4\) years
• Time period 2: \(3\mathrm{n} = 6\) years
• Interest multiplier per year: \((1 + 20/100) = 1.2\)

Step 2: Calculate investment value after 2n = 4 years
Value after 4 years = \(\mathrm{P} \times (1.2)^4 = \$1000 \times 2.0736 = \$2073.60\)

Step 3: Calculate investment value after 3n = 6 years
Value after 6 years = \(\mathrm{P} \times (1.2)^6 = \$1000 \times 2.985984 = \$2985.98\)

Step 4: Find the ratio
How many times greater = Value after 6 years ÷ Value after 4 years
= \(\$2985.98 \div \$2073.60 = 1.44\)

Step 5: Verify with answer choices
We need to check which answer choice gives us \(1.44\) when we substitute our smart numbers:
• Choice C: \((1 + \mathrm{i}/100)^\mathrm{n} = (1 + 20/100)^2 = (1.2)^2 = 1.44\) ✓

Step 6: Confirm the pattern
Notice that \((1.2)^6 \div (1.2)^4 = (1.2)^{6-4} = (1.2)^2 = (1.2)^\mathrm{n}\), confirming our answer.