This section uses a little math, but I’ll go through it slowly in small steps. As with the previous section, this section is not necessary if you don’t want to tackle it.

I want to take a close look at what happens to the payment amount (S) as I take the same amount of loan, at the same interest rate, but for longer and longer times. In other words, I want to look at the payment formula above when I pick values for P, R, and У and leave them alone, but let n get larger and larger.

I’ll introduce the term r for the interest per payment period, just to make things a little neater. That is, let


r = —


so that I can write the payment formula as the much neater looking [10]

If n is very large, then (1 + r)n is much larger than 1, and (1 + r)n – 1 ~ (1 + r)n. This is a fancy way of saying that, for very large values of n, we can ignore the 1 in the denominator of the formula for S above.

This means that, for very large values of n,

S = p, (1 + r) – Pr(1 + r) = Pr.

(1 + r )n -1 (1 + r )n

But this simple value for S (S = Pr) is exactly the value of the first interest payment on the loan. Recall (or go back and look) that I have demonstrated that when your monthly payment is exactly the first interest payment, the balance never changes and you’re just paying interest on the loan—forever. As n gets larger and larger and your payment approaches (slowly falls to) Pr, almost all of your payment goes toward the interest, and you are reducing your balance very, very slowly. This may or may not be a bad thing depending on your needs, your ability to pay, and other opportunities for the money you’re not using to pay this loan. I’ll go into some of these alternatives in later chapters.

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