Compound Interest -Confirming Answer- "When you get a "30 year fixed rate mortgage" on a house, you borrow a certain amount of money at a certain interest rate. You then make the same monthly payment for 30 years. At the end of that time, the loan is fully paid off. The interest on the loan is compounded monthly. Suppose the amount of money you borrow is n dollars, and that the interest rate, compounded monthly, is r%. Find a formula in terms of n and r that gives you the amount of the monthly payment."
I get my answer to be:
$\frac{(n * (1 + r / (100 * 12))^{360})}{360} $
Can anyone confirm that this is correct?

We give a derivation of the correct formula for the monthly payments.  
We will assume that the first monthly payment is made a month after the loan is issued. We also assume that the interest rate $r$ is given as a percentage. Let $P$  be the monthly payment. 
The present value of a payment of $P$ made $k$ months from now is 
$$P\cdot \frac{1}{1+\frac{r}{(100)(12)}}.$$
For brevity, call $1+\frac{r}{(100)(12)}$ by the name $b$.
The present value (PV) of the $360$ payments is
$$P\left(\frac{1}{b}+\frac{1}{b^2}+\frac{1}{b^3}+\cdots+\frac{1}{b^{360}}\right).$$
By the formula for the sum of a finite geometric series, this is
$$P\cdot\frac{b^{360}-1}{(b-1)b^{360}}.$$
Set this equal to $n$ and solve for $P$. We get
$$P=n\cdot (b-1)\frac{b^{360}}{b^{360}-1}.$$
Remark: Your calculation took the initial amount $n$ owed, and calculated what this debt would grow to in $30$ years. Call this amount $H$, for huge. Then you divided $H$ by $360$ to get the monthly payment. 
However, our debt does not remain $n$ for $30$ years. As we make payments, the amount owed decreases. The rate of decrease is quite slow at first, since most of our monthly payments go to interest. But after a while, we are paying significant amounts off the principal each month.
