Chapter 2.3, Problem 14P

A Difference Equation. In this problem, we approach the loan problem in Example $2$ from a discrete view point. This leads to a difference equation rather than a differential equation.

Let $S_0$ be the initial balance of the loan, and let $S_n$ be the balance after $n$ months. Show that

$S_n=\left(1+r\right)S_{n-1}-k$, $n=1,2,3,\dots ,$ (i)

Where $r$ is the monthly interest rate and $k$ is the monthly payment. In Example $2$, the annual interest rate is $8\%$, so here we take $r=0.08/12$.

Let $R=1+r$, so that Eq.(i) becomes

$S_n=R{S}_{n-1}-k$, $n=1,2,3,\dots ,$ (ii)

Find $S_1$, $S_2$, and $S_3$.

Use an induction argument to show that

$S_n=R^n{S}_0-\frac{R^n-1}{R-1}k$ (iii)

For each positive integer $n$.

Let $S_0=20,000$ and suppose that (as in Example $2$) the loan is to be paid off in $48$ months. Find the value of $k$ and compare it with the result of Example $2$.

The result of Example $2$:

$k=\$486.88$

The total amount paid over the life of loan is 48 times $\$486.88$ or $\$23,370.24$.