The formula for finding the amount $A_n$ one has to pay periodically $($let$\text{'}$s just assume it is monthly$),$

over $n$ months to payoff a debt of $P$ dollars and interest rate per period $i$ is given by

$A_n=P\frac{i(1+i{)}^n}{(1+i{)}^n-1}$

This is known as the amortization formula. Note that we may assume that $P>0$ and $$1\mathrm{.}57\%i=\frac{\mathrm{.}07}{12}{A}_{360}=$$$9979\mathrm{.}54A_{n}1+i\frac{1}{1+i}N\frac{1}{1+i}\frac{1}{(1+i{)}^2}nP{\displaystyle \underset{j=1}{\overset{n}{}}}{A}_n\frac{1}{(1+i{)}^j}=P.A_n{A}_n\{1,$dots,$n\}PAAninNAAin(0,1){\displaystyle \underset{j=1}{\overset{n}{}}}{}^j=\frac{1-{}^{n+x^{\text{'}}}}{1-}0.$ For

example consider $a$ $$1\mathrm{.}5$ million mortgage $(about$ the average home price $in$ Santa Cruz$)at7\%$ annual

interest, which corresponds $toi=\frac{\mathrm{.}07}{12}$ monthly interest. $A30$ year, $or360$ month mortgage, would

have a monthly payment $of$ about $A_{360}=$$$9979\mathrm{.}54.$

$(As$ another fun exercise for your own amusement, you can figure out how much you'd have $to$ make $a$

year $to$ pay this mortgage and still have money for income taxes, property taxes, insurance, and maybe

some food.$)$

You will prove that the formula for $A_{n}is$ correct. $Todo$ this, $we$ have $to$ introduce the concept $of$

discounting cash flows, because the bank needs the sum $of$ discounted cash flows from the stream $of$

payments $to$ equal the principle.

A dollar today $is$ worth more than one $in$ the next period, since a dollar today would $be$ worth $1+i$

tomorrow $ifit$ accumulates interest. Equivalently, $we$ can think $of$ one dollar $1$ period from now being

worth $\frac{1}{1+i}$ today. $We$ thus say a dollar $tobe$ received a year from now has $aNet$ Present Value $(NPV)$

$of\frac{1}{1+i}.$ Similarly a dollar $tobe$ collected two periods has $anNPVof\frac{1}{(1+i{)}^2}.$ For our amortization

formula $tobe$ right, the bank needs the stream $of$ cash flows over the $n$ periods $to$ have a total $NPV$

equal the priniciple $P.$ Thus $we$ need

${\displaystyle \underset{j=1}{\overset{n}{}}}{A}_n\frac{1}{(1+i{)}^j}=P.$

$We$ will prove that following proposition.

Proposition: The amount $A_n$ given $in$ equation $(1),$ satisfies equation $(2).$ Thus a stream $of$ payments

$of$ amount $A_n$ each and every month $in$ the set $\{1,$dots,$n\}$ has $anNPVofP.To$ prove this proposition,

$we$ will proceed $in$ the following steps

$(a)$ Use induction $to$ prove that the following Lemma $is$ true:

Lemma AAninN and $AAin(0,1),$

${\displaystyle \underset{j=1}{\overset{n}{}}}{}^j=\frac{1-{}^{n+x^{\text{'}}}}{1-}$