There are hundreds of loan calculators online but none of them tell me this,
Say I take out a 30 year mortgage on a 80,000 dollar house for 4% interest. How much more than $80k will I spend at the end of the 30 years? Assume the normal compounding, etc.
The loan calculator will tell you the monthly payment (or the biweekly payment, if you opt for that).
If the monthly payment is $M$, then to find your total payments, multiply $M$ by $(30)(12)$. Subtract $80000$ to find the answer to your question.
With biweekly payments $B$, to find the total payments we again multiply $B$ by the number of two week periods. This could be a little larger than $(30)(24)$, since the extra $1$ or two days in the year add up to more than a month. Your mortgage document will specify the exact number of payments.
First determine the payment $A$ given the initial principal $G = 80,000$ with yearly interest rate $E = 0.04$ -- so the effective payment (monthly, bi-weekly, etc.) interest rate is $R = 1 + E/W$, where $W$ is the number of payments per year -- over $N = Y W$ payments, where $Y$ is the number of years (typically 15 or 30). Let $P\left(t\right)$ be the principle remaining at the $t$-th payment, so that $P\left(0\right) = G$ and $P\left(N\right) = 0$. Then $$ \begin{eqnarray} P\left(1\right) &=& P\left(0\right) R - A = GR - A \\ P\left(2\right) &=& P\left(1\right) R - A = GR^2 - AR - A \\ &...& \\ P\left(t\right) &=& G R^t - A \sum_{k=0}^{t-1} R^k = G R^t - A\left(\frac{R^t-1}{R-1}\right)\\ \end{eqnarray} $$ Using $P\left(N\right) = 0$: $$ \begin{eqnarray} A = G R^N \left(\frac{R-1}{R^N-1}\right) \end{eqnarray} $$ The total amount paid over the life of the loan is then $AN$, and the total interest paid is $AN-G$. With W = 12 and Y = 30, it looks like the total interest paid is $57,495.61.
You can actually calculate the total interest paid as a fraction of the loan amount $G$: $$ \begin{eqnarray} \frac{AN-G}{G} = \frac{AN}{G} - 1 = N R^N \left(\frac{R-1}{R^N-1}\right) - 1 \end{eqnarray} $$
