Problem
Set 9. Linzee's grandparents, on the day of her Quinceañera ( \( 15^{\text {th }} \) birthday) gave her \( \$ 20,000 \) (even though she missed a sectional basketball game). Linzee having learned about how important it is to start saving early is deciding on two investment options. Option 1 will pay her \( 4.05 \% \) interest compounded annually and Option \( B \) will pay her \( 4.01 \% \) compounded quarterly. a. Write a function for Options 1 and 2 that calculate the value of the account after \( n \) years (1) \( 20,000=(1.0405)^{n} \) (2) \( 20,000=(1.0401 / 4)^{4 n} \) b. Linzee plans leave the money in the account for 10 years and use it to pay for a trip around the world when she finishes college. Which option will yield more money and how much more?
Solution
Step 1 ：\( A_{1}(n) = 20000(1.0405)^{n} \)
Step 2 ：\( A_{2}(n) = 20000(1+ 0.0401/4)^{4n} \)
Step 3 ：\( A_{1}(10) = 20000(1.0405)^{10} \approx 29714.47 \) ; \( A_{2}(10) = 20000(1+ 0.0401/4)^{4 \times 10} \approx 29713.14 \)
Step 4 ：Option 1 will yield more money, the difference is approximately \(29714.47 - 29713.14 \approx \$ 1.33 \)
