For this kind of scenario, let's apply the Quarterly Compounding Formula:
![A\text{ = }P\lbrack(1+r)^(4n)\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/i4djdw5797dtb28meuihqtd3zo68djeyk9.png)
Where,
A = The total balance of the loan including the interest = $8,000.00
P = Principal Amount (Loan amount)
r = Rate of interest = 3%
n = No. of periods = 2 years
Let's plug in the values to the formula. we get,
![\text{ A = }P\lbrack(1+r)^(4n)\rbrack\text{ }\rightarrow\text{ 8,000 = }P\lbrack(1+(3)/(100))^(4(2))\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/zezow7upg41pmj04xjq9xcu7mahxs4mng1.png)
![\text{8,000 = }P\lbrack(1+0.03)^(8))\rbrack\text{ }\rightarrow\text{ 8,000 = }P\lbrack(1.03)^8\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/ivu877ljth3ofvvwj3c61c3hhwzmm6v79t.png)
![\text{8,000 = }P\lbrack(1.03)^8\rbrack\text{ }\rightarrow\text{ 8,000 = P(1.26677008139)}](https://img.qammunity.org/2023/formulas/mathematics/college/pe30rm5huc6hok22zqdzh86mpij2ak431i.png)
![\text{P = }\frac{8,000}{\text{1.26677008139}}\text{ = \$6,315.27}](https://img.qammunity.org/2023/formulas/mathematics/college/wfg6jflrnkmbw1vqeuc3ixutdvpl10vss6.png)
We now get the accumulated interest:
![\text{ Accumulated Interest = A - P = \$8,000 - \$6,315.27}](https://img.qammunity.org/2023/formulas/mathematics/college/l55p40a8xoowtg5ijz29llbq4dn78w8okt.png)
![\text{ Accumulated Interest = \$1,684.73}](https://img.qammunity.org/2023/formulas/mathematics/college/6bc6hwrxpval8hf5fm0xjycvisj7e99scz.png)
Therefore, Aubrey accumulated a total of $1,684.73 interest.