To solve this problem, we can use the compound interest formula:

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

Where:

• $A$ is the amount of money after $t$ years.
• $P$ is the principal amount (initial investment).
• $r$ is the annual interest rate (in decimal form).
• $n$ is the number of times the interest is compounded per year.
• $t$ is the number of years.

Henry's Account:

• Principal $P = 4,100$
• Annual interest rate $r = 3\frac{3}{8}\% = \frac{27}{8}\% = 0.03375$
• Compounded quarterly, so $n = 4$
• Time $t = 20$ years

For Henry, the formula becomes:

$A_H = 4,100 \left(1 + \frac{0.03375}{4}\right)^{4 \times 20}$

Brooklyn's Account:

• Principal $P = 4,100$
• Annual interest rate $r = 2\frac{3}{4}\% = 0.0275$
• Compounded monthly, so $n = 12$
• Time $t = 20$ years

For Brooklyn, the formula becomes:

$A_B = 4,100 \left(1 + \frac{0.0275}{12}\right)^{12 \times 20}$

Now, let's calculate both $A_H$ and $A_B$ and find the difference between them.After 20 years, Henry's account will have approximately $8,030, while Brooklyn's account will have approximately $7,102. The difference between the two accounts is about $928.

Would you like more details on any part of the calculation, or do you have any other questions?

Here are five related questions to explore:

1. How would the difference change if both accounts were compounded annually instead?
2. What if Brooklyn invested $5,000 instead of $4,100?
3. How would a higher interest rate (e.g., 5%) impact the final amounts?
4. What is the total interest earned by each person?
5. How long would it take for Henry's investment to double at his current rate?

Tip: Increasing the frequency of compounding results in slightly more interest, even with the same annual rate.