Guided Project 45: The mathematics of loans
Topics and skills: Sequences and sums
Between houses, cars, or education, most people will take out a loan at some time in their lives. In a typical loan
situation, a person borrows an amount $B_0$ at a fixed interest rate with a fixed payback period. The borrower
makes monthly payments until the loan balance (the amount that remains to be paid) is reduced to zero. The
goal of this project is to set up a mathematical model that describes the balance in the loan for each month. The
results of this project can be calculated and presented very effectively using a spreadsheet.
1. Let's take a specific example. Assume you borrow $B_0 = \$15,000$ with a fixed annual interest rate of 6%, or
0.5% per month. As a first problem, assume that your monthly payment is $500. The goal is to compute the
number of months required to pay off the loan. Every month, two things happen: Interest, which is 0.5% of
the current balance, is added to the current balance and the loan balance is decreased by the monthly
payment of $500. We let $B_n$ be the loan balance after the $n$th payment. Explain why
$B_1 = B_0 + 0.005B_0 - 500 = 1.005B_0 - 500$.
2. Explain why $B_2 = 1.005B_1 - 500$. Show that in general, after the $n$th payment, the balance is
$B_n = 1.005B_{n-1} - 500$, for $n = 1, 2, 3, ....$
3. At this point we have found a recurrence relation for the sequence of loan balances. We now find an
explicit formula for $B_n$. Beginning with the first month, we know $B_1 = 1.005B_0 - 500$. After the second
payment, the loan balance is $B_2 = 1.005^2B_0 - 500(1 + 1.005)$. Show that after the third payment, the loan
balance is $B_3 = 1.005^3B_0 - 500(1 + 1.005 + 1.005^2)$.
4. Show that in general, the loan balance after the $n$th payment is
$B_n = 1.005^nB_0 - 500(1 + 1.005 + 1.005^2 + ... + 1.005^{n-1})$, where $n = 0, 1, 2, ...$
5. We now have an explicit formula for the loan balance after any number of months. Evaluate the geometric
sum in Step 4 and show that $1 + 1.005 + 1.005^2 + ... + 1.005^{n-1} = 200(1.005^n - 1)$.
