How to explain mortgage monthly payment formula using school math?

Question

$P = L*\frac{x*(1+x)^n}{(1+x)^n - 1}$

where

  P - monthly payment

  L - loan amount

  x - monthly interest rate  

  n - number of payments 

I answer my own question.

Answer 1

OK. Let me state the problem here.

We have a loan amount $L$, number of payments $n$, interest rate per year $Y$. What will a monthly payment $P$ be?

Usually loan has interest rate per year $Y$ and payments made by month, so we need to convert $Y$ to monthly interest rate $x = \frac{Y}{12} (1)$.

Let's calculate the loan reminder after the 1st payment, we used all loan $L$ money for a month.

$L_1 = L*(1+x) - P (2)$

Let's calculate the loan reminder after the 2nd payment, we used the loan after the 1st payment $L_1$ money for a month. Use formula (2) to simplify and to open parenthesis.

$L_2=L_1*(1+x) - P = [L*(1+x) - P]*(1+x) - P = L*(1+x)^2 -P*(1+x) -P = L*(1+x)^2 -P*[(1+x) + 1] (3)$

Let's calculate the loan reminder after the 3rd payment, we used the loan after the 2nd payment $L_2$ money for a month. Use formula (3) to simplify and to open parenthesis.

$L_3=L_2*(1+x) - P = \{L*(1+x)^2 -P*[(1+x) + 1]\}*(1+x) - P = L*(1+x)^3 -P*(1+x)^2 - P*(1+x) -P = L*(1+x)^3 -P*[(1+x)^2 + (1+x) + 1] (4)$

We can see a pattern now and can write the expression for $L_n$.

$L_n=L_{n-1}*(1+x) - P = L*(1+x)^n - P*[(1+x)^{n-1}+ ... + (1+x)^2 + (1+x) + 1] (5)$

I also want to state that after last payment $L_n$ we are no longer own any amount on the loan. Or $L_n = 0 (6)$ or using (5) in (6)

$L*(1+x)^n - P*[(1+x)^{n-1}+ ... + (1+x)^2 + (1+x) + 1] = 0(6')$

Let's simplify $(1+x)^{n-1}+ ... + (1+x)^2 + (1+x) + 1$. We know from school mathematical curriculum about geometric progression. Each element of geometric progression can be expressed as

$a_k=a*r^{k-1}$ (6")

where $1 \le k \le K$ and $r$ is the denominator of the geometric progression.

and the sum of elements of geometric progression from $1$ to $k$ is calculated by formula

$S_k = a*\frac{1-r^k}{1-r}$ (6''')

It can be seen that $a=1$ and $r=(1+x)$, so $(1+x)^0, (1+x)^1, (1+x)^2, ..., (1+x)^{n-1}$. Let's apply the formula (6''') to our case: $(1+x)^{n-1}+ ... + (1+x)^2 + (1+x) + 1= 1* \frac{1-(1+x)^n}{1-(1+x)} = \frac{1-(1+x)^n}{1-1-x} = -\frac{1-(1+x)^n}{x}(7)$.

Now use (7) in (6')

$L*(1+x)^n - P* [-\frac{1-(1+x)^n}{x}]= 0 (8)$

Solving (8) to find $P$

$L*(1+x)^n = P* [-\frac{1-(1+x)^n}{x}] (8')$

$L*(1+x)^n = P* [\frac{(1+x)^n - 1}{x}] (8'')$

$P = L*[\frac{x*(1+x)^n}{(1+x)^n-1}](9)$

Here's the small piece of Python code to calculate monthly payment:

Y=float(input("Enter the yearly interest rate (in percent):"))
n=int(input("Enter the number of payments:"))
L=float(input("Loan amount?"))

x=Y/(12*100)
r=1+x
r1=r**n
P=L*x*r1/(r1-1)
print("The monthly payment ", P)

Test results:

Enter the yearly interest rate (in percent):4
Enter the number of payments:360
Loan amount?100000
The monthly payment  477.4152954654538

Questions, comments, edits ?

Answer 2

Here is a second approach:

Let the interest rate per month be $x$ and the number of payments be $n$.

Just assume that each month $P$ is paid back to be bank but never discounted from the loan. They are just stored in the bank and create compound interests by themselves.

After $n$ instalments, the total sum of the amounts created by the the $n$ instalments is equal to the amount created by the loan $L$ after a period of $n$ instalments.

The debt is then settled directly.

Not sure whether I explain the concepts involved clearly. Hope that the following steps below help.

For the first $P$, after another $n-1$ payments, it amounts to $P(1+x)^{n-1}$.

For the second $P$, after another $n-2$ payments, it amounts to $P(1+x)^{n-2}$.

For the third $P$, it amounts to $P(1+x)^{n-2}$.

...

For the last but one $P$, it amounts to $P(1+x)$.

For the last $P$, it has no interest. Therefore it is just $P$.

Thus the total amount is $$P(1+x)^{n-1}+P(1+x)^{n-2}+ \cdots + P(1+x)+P=P\left(\frac{(1+x)^n-1}{x}\right) \tag{1}$$

On the other hand, after $n$ payments, the total amount created by the loan $L$ is $$L(1+x)^n \tag{2}$$

For a fair play, the amounts in $(1)$ and $(2)$ equate.

$$\therefore P\left(\frac{(1+x)^n-1}{x}\right)=L(1+x)^n$$ $$P=\frac{Lx(1+x)^n}{(1+x)^n-1}$$