Question 949625
{{{"9%"=9/100=0.09}}} nominal rate convertible monthly means that
each {{{1/12}}} of the year (1 month) they add  {{{1/12}}} of {{{"9%"}}} ,
that is {{{(1/12)*0.09=0.0075}}} times the old balance,
to the old balance to calculate the new balance.
 
At the end of the first month, just as Mark is about to make his first payment,
they add {{{0.0075*("$40,000")}}} to the {{{"$40,000"}}} to get 
{{{"$40,000"+0.0075*("$40,000")=(1+0.0075)*("$40,000")=1.0075*("$40,000")}}} ,
which is {{{1.0075}}} times the previous balance.
Immediately, Mark makes his {{{"$735"}}} payment, and the balance goes down to
{{{1.0075*("$40,000")-"$735"}}} .
 
At the end of the second month, just as Mark is about to make his second payment,
the balance again increases by a factor of {{{1.0075}}} ,
from {{{1.0075*("$40,000")-"$735"}}} to {{{1.0075^2*("$40,000")-"$735"*1.0075}}} ,
only to get immediately reduced by the second {{{"$735"}}} payment
to {{{1.0075^2*("$40,000")-"$735"*1.0075-"$735"}}} .
 
At the end of the second month,
the balance first increases again by a factor of {{{1.0075}}} ,
to {{{1.0075^3*("$40,000")-"$735"*1.0075^2-"$735"*1.0075}}} , and then
decreases to {{{1.0075^3*("$40,000")-"$735"*1.0075^2-"$735"*1.0075-"$735"}}}
with the third {{{"$735"}}} payment.
 
The pattern repeats each month, so after {{{12*5=60}}} months, and {{{60}}} payments, the balance is
{{{1.0075^60*("$40,000")-"$735"*1.0075^59-"$735"*1.0075^58-"..."-"$735"*1.0075^2-"$735"*1.0075-"$735"}}}
={{{1.0075^60*("$40,000")-"$735"*(1.0075^59+1.0075^58+"..."+1.0075^2+1.0075+1)}}}
From either geometric sequences and series,
or from polynomial s and factoring, we know that
{{{1.0075^59+1.0075^58+"..."+1.0075^2+1.0075+1=(1.0075^60-1)/(1.0075-1)}}} .
So the final balance, after {{{60}}} months, and {{{60}}} payments is
{{{1.0075^60*("$40,000")-"$735"*(1.0075^60-1)/(1.0075-1)=highlight("$7,190.50")}}}