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 Question from Kenneth: I have a question regarding the loan payment formula shown below. Calculating the Payment Amount per Period The formula for calculating the payment amount is shown below. Simple Amortization Calculation Formula A = P X r(1 + r)n over (1 + r)n - 1 where A = payment Amount per period P = initial Principal (loan amount) r = interest rate per period n = total number of payments or periods Is this formula/calculation a condensed version of a longer calculation? I am curious to know how the (1 +r)n - 1 was developed from the longer calculation. For example, r(1 + r)n may have been (r + rn)n. The n's are exponents. I thank you for whatever helpful explanation that may be provided. Kenneth

Hi Kenneth,

First of all you need to know how to calculate the sum of a finite geometric series. If you have a geometric sequence, $a, at , at^2, at^3, \cdot, \cdot, \cdot$ and $S_n$ is the sum of the first $n$ terms of this sequence then

$S_n = a + at + at^2 + at^3 + \cdot, \cdot \cdot + at^{n-1} = \frac{a \left( 1 - t^n\right)} {1-t}.$

A proof appears in Penny's response to a previous question. I have changed Penny's notation as you have already used the variable $r.$ We will use this result later.

You have a loan amount of $P$ dollars nd an interest rate of $p$ per period. At the end of each period you make a payment of $A$ dollars. At the end of the first period the amount you owe has grown to $P(1+r)$ dollars and you pay

$A$ dollars so at the beginning of the second period you owe

$P(1+r) - A \mbox{ dollars.}$

At the end of the second period the amount you owe is $(P(1+r) - A)(1+r)$ dollars and you pay $A$ dollars so at the beginning of the third period you owe

$(P(1+r) - A)(1+r) -A = P(1+r)^2 - A\left( (1+r) + 1 \right) \mbox{ dollars.}$

Continuing in this fashion at the end of the third period you owe

$P(1+r)^3 - A\left( (1 + r)^2 + (1+r) + 1 \right) \mbox{ dollars}$

and at the end of the $n^{th}$ period you owe

$P(1+r)^n - A\left( (1+r)^{n-1} + \cdot \cdot \cdot + (1 + r)^2 + (1+r) + 1 \right) \mbox{ dollars}$

But at the end of the $n^{th}$ period you have paid off the loan so

$P(1+r)^n - A\left( (1+r)^{n-1} + \cdot \cdot \cdot + (1 + r)^2 + (1+r) + 1 \right) = 0.$

At this point you ned to use the expression for the sum of the first $n$ terms of a geometric sequence given above with $t = 1+r$ which gives

$P(1+r)^n - A\left( \frac{1 - (1+r)^n}{1 - (1+r)} \right) = 0.$

Solving this equation for $A$ gives

$A = P \left( \frac{r(1+r)^n}{(1+r)^n - 1}\right) \mbox{ dollars.}$

Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.