Let's break down and compare the growth of Person A's and Person B's investments over time, using the formula for compound interest:

1. Compound Interest Formula:

The compound interest formula is:

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

Where:

• $A$ is the amount of money accumulated after $t$ years, including interest.
• $P$ is the principal amount (the initial deposit).
• $r$ is the annual interest rate (in decimal form).
• $n$ is the number of times interest is compounded per year.
• $t$ is the time the money is invested for in years.

2. For Person A:

• Principal $P_A = 2600$
• Annual interest rate $r_A = 6\% = 0.06$
• Compounded annually ($n_A = 1$)
• The formula for Person A simplifies to:

$A_A = 2600 \left(1 + 0.06 \right)^t = 2600 \times 1.06^t$

3. For Person B:

• Principal $P_B = 2400$
• Annual interest rate $r_B = 7\% = 0.07$
• Compounded monthly ($n_B = 12$)
• The formula for Person B becomes:

$A_B = 2400 \left(1 + \frac{0.07}{12} \right)^{12t} = 2400 \left(1 + 0.005833 \right)^{12t}$

Now, these equations can be used to calculate the future value of both investments after any number of years $t$.

Would you like to see the results for a specific number of years or further compare their growth rates? Let me know how you'd like to proceed!

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Here are 5 related questions to expand your understanding:

1. How does the frequency of compounding affect the total amount of interest earned?
2. How does a small difference in interest rates (6% vs. 7%) affect long-term growth?
3. What is the difference in total interest earned if both amounts are left in the accounts for 10 years?
4. How does increasing the number of compounding periods affect Person B's investment?
5. What would happen if both people increased their principal deposits by 10%?

Tip: The more frequent the compounding, the faster the investment grows due to interest earning on interest more often.