Question

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.

Answer 1

`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")`

Answer 2

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`