Question

A debt of $10,000 is to be amortized by equal payments of $400 at the end of each month, plus a final payment after the last $400 payment is made. If the interest is at the rate of 1% compounded monthly (the same as an annual rate of 12% compounded monthly), i. Write a discrete dynamical system that models the situation. ii. Construct a table showing the amortization schedule for the required payments. iii. Find a solution for the system.

Answer 1

\[A _{n}=A _{n-1}(1.01)-400\]

Answer 2

is that correct?

Answer 3

or \[A _{n}=10000(1.01)^{n-1}-400\]

Answer 4

\[A _{n}=A _{n-1}(1.01)-400\]

Answer 5

An=An-1(1.01(-400

Answer 6

so my last equation was right?

Answer 7

Here is how I would solve
P-Principle
D-Monthly Payment
R-(Monthly interest rate)


1st Month - P(1+R)-D
2nd Month - [P(1+R)-D][(1+R)]-D =\[P(1+R)^2 - D(1+R)-D=P(1+R)^2 - D(1+(1+R))\]
3rd Month \[P(1+R)3−D(1+(1+R)+(1+R)^2)\]

Nth Month - \[P(1+R)N−D(1+(1+R)+(1+R)^2+.....+(1+R)^{N-1})\]

\[(1+(1+R)+(1+R)^2+.....+(1+R)^{N-1})\] is a gemotric series

Formula For Sum of Geomtric Series
\[\sum_{0}^{N}v^n={( 1-v^{n+1})} /(1-v)\]

our equation is :
\[P(1+R)^N-D\left(-\frac{1-(1+R)^N}{R}\right)\]
Simplified to
\[\frac{D \left(1-(R+1)^N\right)}{R}+P (R+1)^N\]

For this particular problem:
\[\frac{400\left(1-(1.01)^N\right)}{.01}+10000(1.01)^N=0\]
N=28.91 Month

Answer 8

I don't think my uni has any 1 credit hour class