If you have not been given a formula for the monthly payment amount, this can be derived as follows:
Let P = monthly payment, A = amount borrowed, i = monthly interest rate, and n = the number of payments.
Also, let \(\displaystyle D_n\) be the debt amount after payment n.
Consider the recursion:
(1) \(\displaystyle D_{n}=(1+i)D_{n-1}-P\)
(2) \(\displaystyle D_{n+1}=(1+i)D_{n}-P\)
Subtracting (1) from (2) yields the homogeneous recursion:
\(\displaystyle D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}\)
whose associated auxiliary equation is:
\(\displaystyle r^2-(2+i)r+(1+i)=0\)
\(\displaystyle (r-(1+i))(r-1)=0\)
Thus, the closed-form for our recursion is:
\(\displaystyle D_n=k_1(1+i)^n+k_2\)
Using initial values, we may determine the parameters \(\displaystyle k_i\):
\(\displaystyle D_0=k_1+k_2=A\)
\(\displaystyle D_1=k_1(1+i)+k_2=(1+i)A-P\)
Solving this system, we find:
\(\displaystyle k_1=\dfrac{Ai-P}{i},\,k_2=\dfrac{P}{i}\) and so we have:
\(\displaystyle D_n=\left(\dfrac{Ai-P}{i} \right)(1+i)^n+\left(\dfrac{P}{i} \right)=\dfrac{(Ai-P)(1+i)^n+P}{i}\)
Now, equating this to zero, we can solve for P:
\(\displaystyle \dfrac{(Ai-P)(1+i)^n+P}{i}=0\)
\(\displaystyle (Ai-P)(1+i)^n+P=0\)
\(\displaystyle (P-Ai)(1+i)^n=P\)
\(\displaystyle P\left((1+i)^n-1 \right)=Ai(1+i)^n\)
\(\displaystyle P=\dfrac{Ai(1+i)^n}{(1+i)^n-1}\)
\(\displaystyle P=\dfrac{Ai}{1-(1+i)^{-n}}\)
Now, can you determine what their monthly payment amount is?
