Construct an equation that models the repayment of a high value loan. Such as a mortgage. The loan is based on compound interest with a monthly calculation and repayment cycle. The site system forcing me to add a question mark: ?

Set the monthly repayment as R, the annual percent interest (apr) as T% and the initial principle sum as #P_0#

This question is set so that I may demonstrate some mathematical processes.

2 Answers
Tony B
Feb 11, 2017

#color(blue)("Solution part 1")#

#color(red)("With full explanation this is a big solution so I am splitting it")#

#ul("Starting point")#

Let the number of years be #y#
Let the count of calculation cycles be #n# so #n=12y#

Given that the initial principle sum is #P_0#
Set the adjusted principle after the 1st cycle as #P_1#
Set the adjusted principle after the 2nd cycle as #P_2#
Set the adjusted principle after the 3rd cycle as #P_3# and so on

The interest for 1 year is #T%->T/100#
So splitting this over each month gives #T%/12->T/1200#

#color(brown)("First payment cycle")#

#P_1=P_o(1+T/1200)-R#

#color(brown)("Second payment cycle")#

#P_2=P_1(1+T/1200)-R#

#color(brown)("Third payment cycle")#

#P_3=P_2(1+T/1200)-R#

#color(brown)("Fourth payment cycle")#

#P_4=P_3(1+T/1200)-R#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Repeating this process but with full substitution

#color(brown)("Second payment cycle")#

#P_2=color(green)([P_o(1+T/1200)-R] color(purple)((1+T/1200)-R)#

#P_2=P_0(1+T/1200)^2-R(1+T/1200)-R#
..........................................................................
#color(brown)("Third payment cycle")#

#P_3=#

#color(green)([P_0(1+T/1200)^2-R(1+T/1200)-R]color(purple)((1+T/1200)-R#

#P_3=P_0(1+T/12)^3-R(1+T/1200)^2-R(1+T/100)-R#
..........................................................................

#color(brown)("Fourth payment cycle")#

Using the same approach we end up with:

#P_4=P_0(1+T/12)^4-R(1+T/1200)^3-R(1+T/1200)^2-R(1+T/1200)-R#
..........................................................................
Let #x=(1+T/1200)# giving:

#P_4=P_0x^4-Rx^3-Rx^2-Rx-R#

Factor out the #-R# giving:

#P_4=P_0x^4-R(x^3+x^2+x+1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From this it is obvious that ( hate that phrase!)
For any n we have:

#P_n=P_0x^n-R(x^(n-1)+x^(n-2)+x^(n-3)+...+x+1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(purple)("See part 2 next")#

Tony B
Feb 11, 2017

With full explanation this is a big solution so I am splitting it

See Solution part 1 first

#color(blue)("Solution part 2")#

Following on from:

Set as #Equation(1)#
#P_n=P_0x^n-R(x^(n-1)+x^(n-2)+x^(n-3)+...+x+1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that at the completion of paying off the loan #P_n=0#

To take this further we need to determine the sum of the series within the brackets.

set
#s=x^(n-1)+x^(n-2)+x^(n-3)+...+x+1" ".......Equation(2)#

Then
#sx=x^n+x^(n-1)+x^(n-2)+...+x^2+x" " .......Equation(3)#

#Equation(3)-Equation(2)# gives:

#s=cancel(x^(n-1))+cancel(x^(n-2))+cancel(x^(n-3))+...+cancel(x)+1" ".......Equation(2)#
#sx=x^n+cancel(x^(n-1))+cancel(x^(n-2))+...+cancel(x^2)+cancel(x)" " .......Equation(3)#

#sx-s=x^n-1#

Factor out the #s#

#s(x-1)=x^n-1#

#s=(x^n-1)/(x-1) " "..............Equation(4)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute #Equation(4)# into #Equation(1)# giving:

#P_n=P_0x^n-Rs#

#P_n=P_0x^n-(R(x^(n-1)-1))/(x-1)#

But #x=(1+T/1200)# giving:

#P_n=P_0(1+T/1200)^n-(R[(1+T/1200)^(n-1)-1])/((1+T/1200)-1)#

#P_n=P_0(1+T/1200)^n- (1200R)/T[ (1 +T/1200)^(n-1)-1]#

To determine the different values set #P_n=0#

Do not forget that #n# is months so the year count #y# is such that:

#y=n/12#

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