To solve this problem, we compare the future values of an investment under quarterly compounding versus annual compounding. We use the compound interest formula:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Where:

• $A$ = future value of the investment
• $P$ = principal amount (initial investment) = 2000
• $r$ = annual interest rate = 5% or 0.05
• $n$ = number of compounding periods per year
• $t$ = time (in years) = 10

1. Future Value with Annual Compounding

For annual compounding, $n = 1$: $A_{\text{annual}} = P \left(1 + \frac{r}{n}\right)^{nt}$ $A_{\text{annual}} = 2000 \left(1 + \frac{0.05}{1}\right)^{1 \times 10}$ $A_{\text{annual}} = 2000 \left(1.05\right)^{10}$ Using a calculator: $A_{\text{annual}} \approx 2000 \times 1.62889 \approx 3257.78$

2. Future Value with Quarterly Compounding

For quarterly compounding, $n = 4$: $A_{\text{quarterly}} = P \left(1 + \frac{r}{n}\right)^{nt}$ $A_{\text{quarterly}} = 2000 \left(1 + \frac{0.05}{4}\right)^{4 \times 10}$ $A_{\text{quarterly}} = 2000 \left(1 + 0.0125\right)^{40}$ $A_{\text{quarterly}} = 2000 \left(1.0125\right)^{40}$ Using a calculator for $(1.0125)^{40}$: $(1.0125)^{40} \approx 1.64362$ $A_{\text{quarterly}} \approx 2000 \times 1.64362 \approx 3287.24$

3. Difference in Values

Now, subtract the future values: $\text{Difference} = A_{\text{quarterly}} - A_{\text{annual}}$ $\text{Difference} \approx 3287.24 - 3257.78 \approx 29.46$

Final Answer:

The difference in value after 10 years is approximately $29.46.

Would you like me to explain any part in further detail?

Related Questions:

1. How is the compound interest formula derived?
2. How does compounding frequency affect future value growth?
3. What happens to the investment value if the interest is compounded monthly?
4. What if the interest rate changes over time—how is that calculated?
5. How does simple interest differ from compound interest?

Tip: More frequent compounding generally increases the future value of an investment due to the accumulation of interest on previously earned interest.