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

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

Where:

• $A$ = the future value of the investment/loan, including interest.
• $P$ = the principal investment amount (the initial deposit or loan amount).
• $r$ = the annual interest rate (decimal).
• $n$ = the number of times that interest is compounded per year.
• $t$ = the time the money is invested or borrowed for, in years.

Given:

• $P = 6000$
• $r = 8\% = 0.08$
• $n = 12$ (compounded monthly)
• $t = 21$ years

Plugging in the values:

$A = 6000 \left(1 + \frac{0.08}{12}\right)^{12 \times 21}$

Simplifying:

$A = 6000 \left(1 + 0.00667\right)^{252}$

$A = 6000 \left(1.00667\right)^{252}$

Now, let's calculate $A$:

$A \approx 6000 \times 4.9189 = 29,513.40$

The value of the account at the child’s twenty-first birthday will be approximately $29,513.40.

Would you like more details on any step of this process or have any questions?

Related questions:

1. What if the interest rate was compounded quarterly instead of monthly?
2. How would the final value change if the interest rate were 6% instead of 8%?
3. What is the formula for simple interest, and how does it compare to compound interest?
4. How much would the final value be if the deposit were $10,000 instead of $6,000?
5. How much total interest would be earned by the time the child turns 21?

Tip:

When interest is compounded more frequently, the final amount grows larger because interest is calculated on previously earned interest more often.