Questions: At the end of one year, how does the FV with semi-annual compounding compare with the FV with annual compounding? It will be lower It will be the same It will be greater

Solution Steps

To compare the future value (FV) with semi-annual compounding to the FV with annual compounding, we need to calculate both values using the same principal, interest rate, and time period. The formula for FV with compounding is \( FV = P \times (1 + \frac{r}{n})^{nt} \), where \( P \) is the principal, \( r \) is the annual interest rate, \( n \) is the number of compounding periods per year, and \( t \) is the time in years. For semi-annual compounding, \( n = 2 \), and for annual compounding, \( n = 1 \). We will compare the two results to determine which is greater.

Let \( P = 1000 \) be the principal amount, \( r = 0.05 \) be the annual interest rate, and \( t = 1 \) be the time period in years.

Using the formula for future value with compounding, we have: \[ FV_{\text{semi-annual}} = P \times \left(1 + \frac{r}{n_{\text{semi-annual}}}\right)^{n_{\text{semi-annual}} \cdot t} \] where \( n_{\text{semi-annual}} = 2 \). Substituting the values: \[ FV_{\text{semi-annual}} = 1000 \times \left(1 + \frac{0.05}{2}\right)^{2 \cdot 1} = 1000 \times (1 + 0.025)^{2} = 1000 \times (1.025)^{2} \approx 1050.625 \]

Similarly, for annual compounding: \[ FV_{\text{annual}} = P \times \left(1 + \frac{r}{n_{\text{annual}}}\right)^{n_{\text{annual}} \cdot t} \] where \( n_{\text{annual}} = 1 \). Substituting the values: \[ FV_{\text{annual}} = 1000 \times \left(1 + \frac{0.05}{1}\right)^{1 \cdot 1} = 1000 \times (1 + 0.05)^{1} = 1000 \times 1.05 = 1050.0 \]

Now we compare the two future values: \[ FV_{\text{semi-annual}} \approx 1050.625 \quad \text{and} \quad FV_{\text{annual}} = 1050.0 \] Since \( 1050.625 > 1050.0 \), we conclude that the future value with semi-annual compounding is greater than that with annual compounding.

Final Answer