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Sagot :
To find the interest earned for an investment of [tex]$10,000 at an annual interest rate of 14%, compounded quarterly over 13 years, we need to use the compound interest formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial sum of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times the interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed for.
Given:
- \( P = 10,000 \)
- \( r = 0.14 \) (14% as a decimal)
- \( n = 4 \) (since the interest is compounded quarterly)
- \( t = 13 \)
First, we need to calculate the accumulated amount \( A \):
\[ A = 10,000 \left(1 + \frac{0.14}{4}\right)^{4 \times 13} \]
Step 1: Calculate the quarterly interest rate:
\[ \frac{0.14}{4} = 0.035 \]
Step 2: Calculate the exponent \( 4 \times 13 \):
\[ 4 \times 13 = 52 \]
Step 3: Add 1 to the quarterly interest rate:
\[ 1 + 0.035 = 1.035 \]
Step 4: Raise this to the power of 52:
\[ 1.035^{52} \]
Step 5: Multiply by the principal amount:
\[ 10,000 \times 1.035^{52} \]
Using these computations, we find:
\[ A = 10,000 \left(1.035^{52}\right) \]
After calculating the precise value, we subtract the principal amount to find the interest earned:
\[ \text{Interest Earned} = A - P \]
Thus,
\[ \text{Interest Earned} = 10,000 \left(1.035^{52}\right) - 10,000 \]
Finally, we get the result:
The interest earned in 13 years is \$[/tex]49,827.13 (rounded to two decimal places).
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