## anonymous 5 years ago 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.

1. anonymous

$A _{n}=A _{n-1}(1.01)-400$

2. anonymous

is that correct?

3. anonymous

or $A _{n}=10000(1.01)^{n-1}-400$

4. anonymous

$A _{n}=A _{n-1}(1.01)-400$

5. anonymous

An=An-1(1.01(-400

6. anonymous

so my last equation was right?

7. anonymous

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

8. anonymous

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