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\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & Principal & Interest rate & \begin{tabular}{l}
Compounding \end{tabular} & \begin{tabular}{l}
Number of times \\ compounded per year \end{tabular} & Term (Years) & Future Value \\
\hline 2 & [tex]$\$[/tex] 10,000.00[tex]$ & 3.5 & Annually & 1 & 3 & \\
\hline 3 & $[/tex]\[tex]$ 10,000.00$[/tex] & 3.5 & Quarterly & 4 & 3 & \\
\hline 4 & [tex]$\$[/tex] 10,000.00[tex]$ & 3.5 & Monthly & 12 & 3 & \\
\hline 5 & $[/tex]\[tex]$ 10,000.00$[/tex] & 3.5 & Weekly & 52 & 3 & \\
\hline 6 & [tex]$\$[/tex] 10,000.00[tex]$ & 3.5 & Daily & 365 & 3 & \\
\hline 7 & $[/tex]\[tex]$ 10,000.00$[/tex] & 3.5 & Hourly & 8760 & 3 & \\
\hline
\end{tabular}

1. Enter a principal you'd like to invest in cell A2. This value will automatically be copied to the other cells in column A.
2. Enter an interest rate you think you'll earn in cell B2. (Don't forget to enter your percent as a decimal.) This value will automatically be copied to the other cells in column B.
3. Complete column D with the correct number of times interest will be compounded based on the description in column C.
4. Enter a term in years for your investment in cell E2. This value will automatically be copied to the other cells in column E.
5. Create a formula that will calculate the future value in cell F2. Then use the fill-down feature to calculate the remaining values in column F.

Sagot :

GinnyAnswer
To determine the future value of an investment with different compounding frequencies, we will use the compound interest formula:

[tex]\[ FV = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]

Where:
- [tex]\( FV \)[/tex] is the future value of the investment.
- [tex]\( P \)[/tex] is the principal amount (initial investment).
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal).
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested.

### Given Data
- Principal Amount ([tex]\( P \)[/tex]): \[tex]$10,000.00 - Annual Interest Rate (\( r \)): 3.5% (or 0.035 as a decimal) - Term (\( t \)): 3 years - Compounding Frequencies: - Annually: \( n = 1 \) - Quarterly: \( n = 4 \) - Monthly: \( n = 12 \) - Weekly: \( n = 52 \) - Daily: \( n = 365 \) - Hourly: \( n = 8760 \) ### Step-by-Step Calculations #### 1. Annually (n = 1) \[ FV = 10000 \left(1 + \frac{0.035}{1}\right)^{1 \cdot 3} \] \[ FV \approx \$[/tex]11087.18 \]

#### 2. Quarterly (n = 4)
[tex]\[ FV = 10000 \left(1 + \frac{0.035}{4}\right)^{4 \cdot 3} \][/tex]
[tex]\[ FV \approx \$11102.03 \][/tex]

#### 3. Monthly (n = 12)
[tex]\[ FV = 10000 \left(1 + \frac{0.035}{12}\right)^{12 \cdot 3} \][/tex]
[tex]\[ FV \approx \$11105.41 \][/tex]

#### 4. Weekly (n = 52)
[tex]\[ FV = 10000 \left(1 + \frac{0.035}{52}\right)^{52 \cdot 3} \][/tex]
[tex]\[ FV \approx \$11106.71 \][/tex]

#### 5. Daily (n = 365)
[tex]\[ FV = 10000 \left(1 + \frac{0.035}{365}\right)^{365 \cdot 3} \][/tex]
[tex]\[ FV \approx \$11107.05 \][/tex]

#### 6. Hourly (n = 8760)
[tex]\[ FV = 10000 \left(1 + \frac{0.035}{8760}\right)^{8760 \cdot 3} \][/tex]
[tex]\[ FV \approx \$11107.10 \][/tex]

### Summary of Results

- Annually: \[tex]$11087.18 - Quarterly: \$[/tex]11102.03
- Monthly: \[tex]$11105.41 - Weekly: \$[/tex]11106.71
- Daily: \[tex]$11107.05 - Hourly: \$[/tex]11107.10

These values represent the future value of a \$10,000 investment over 3 years at an annual interest rate of 3.5%, compounded at different frequencies. Notice that more frequent compounding results in slightly higher future values. Fill in the corresponding results in the table under the "Future Value" column.
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