# 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.

Feb 11, 2017

$\textcolor{b l u e}{\text{Solution part 1}}$

$\textcolor{red}{\text{With full explanation this is a big solution so I am splitting it}}$

$\underline{\text{Starting point}}$

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

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

$\textcolor{b r o w n}{\text{First payment cycle}}$

${P}_{1} = {P}_{o} \left(1 + \frac{T}{1200}\right) - R$

$\textcolor{b r o w n}{\text{Second payment cycle}}$

${P}_{2} = {P}_{1} \left(1 + \frac{T}{1200}\right) - R$

$\textcolor{b r o w n}{\text{Third payment cycle}}$

${P}_{3} = {P}_{2} \left(1 + \frac{T}{1200}\right) - R$

$\textcolor{b r o w n}{\text{Fourth payment cycle}}$

${P}_{4} = {P}_{3} \left(1 + \frac{T}{1200}\right) - R$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Repeating this process but with full substitution

$\textcolor{b r o w n}{\text{Second payment cycle}}$

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

${P}_{2} = {P}_{0} {\left(1 + \frac{T}{1200}\right)}^{2} - R \left(1 + \frac{T}{1200}\right) - R$
..........................................................................
$\textcolor{b r o w n}{\text{Third payment cycle}}$

${P}_{3} =$

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

${P}_{3} = {P}_{0} {\left(1 + \frac{T}{12}\right)}^{3} - R {\left(1 + \frac{T}{1200}\right)}^{2} - R \left(1 + \frac{T}{100}\right) - R$
..........................................................................

$\textcolor{b r o w n}{\text{Fourth payment cycle}}$

Using the same approach we end up with:

${P}_{4} = {P}_{0} {\left(1 + \frac{T}{12}\right)}^{4} - R {\left(1 + \frac{T}{1200}\right)}^{3} - R {\left(1 + \frac{T}{1200}\right)}^{2} - R \left(1 + \frac{T}{1200}\right) - R$
..........................................................................
Let $x = \left(1 + \frac{T}{1200}\right)$ giving:

${P}_{4} = {P}_{0} {x}^{4} - R {x}^{3} - R {x}^{2} - R x - R$

Factor out the $- R$ giving:

${P}_{4} = {P}_{0} {x}^{4} - R \left({x}^{3} + {x}^{2} + x + 1\right)$

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

${P}_{n} = {P}_{0} {x}^{n} - R \left({x}^{n - 1} + {x}^{n - 2} + {x}^{n - 3} + \ldots + x + 1\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{p u r p \le}{\text{See part 2 next}}$

Feb 11, 2017

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

See Solution part 1 first

$\textcolor{b l u e}{\text{Solution part 2}}$

Following on from:

Set as $E q u a t i o n \left(1\right)$
${P}_{n} = {P}_{0} {x}^{n} - R \left({x}^{n - 1} + {x}^{n - 2} + {x}^{n - 3} + \ldots + x + 1\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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
$s = {x}^{n - 1} + {x}^{n - 2} + {x}^{n - 3} + \ldots + x + 1 \text{ } \ldots \ldots . E q u a t i o n \left(2\right)$

Then
$s x = {x}^{n} + {x}^{n - 1} + {x}^{n - 2} + \ldots + {x}^{2} + x \text{ } \ldots \ldots . E q u a t i o n \left(3\right)$

$E q u a t i o n \left(3\right) - E q u a t i o n \left(2\right)$ gives:

$s = \cancel{{x}^{n - 1}} + \cancel{{x}^{n - 2}} + \cancel{{x}^{n - 3}} + \ldots + \cancel{x} + 1 \text{ } \ldots \ldots . E q u a t i o n \left(2\right)$
$s x = {x}^{n} + \cancel{{x}^{n - 1}} + \cancel{{x}^{n - 2}} + \ldots + \cancel{{x}^{2}} + \cancel{x} \text{ } \ldots \ldots . E q u a t i o n \left(3\right)$

$s x - s = {x}^{n} - 1$

Factor out the $s$

$s \left(x - 1\right) = {x}^{n} - 1$

$s = \frac{{x}^{n} - 1}{x - 1} \text{ } \ldots \ldots \ldots \ldots . . E q u a t i o n \left(4\right)$

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

Substitute $E q u a t i o n \left(4\right)$ into $E q u a t i o n \left(1\right)$ giving:

${P}_{n} = {P}_{0} {x}^{n} - R s$

${P}_{n} = {P}_{0} {x}^{n} - \frac{R \left({x}^{n - 1} - 1\right)}{x - 1}$

But $x = \left(1 + \frac{T}{1200}\right)$ giving:

${P}_{n} = {P}_{0} {\left(1 + \frac{T}{1200}\right)}^{n} - \frac{R \left[{\left(1 + \frac{T}{1200}\right)}^{n - 1} - 1\right]}{\left(1 + \frac{T}{1200}\right) - 1}$

${P}_{n} = {P}_{0} {\left(1 + \frac{T}{1200}\right)}^{n} - \frac{1200 R}{T} \left[{\left(1 + \frac{T}{1200}\right)}^{n - 1} - 1\right]$

To determine the different values set ${P}_{n} = 0$

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

$y = \frac{n}{12}$