Mortgages

The formula for the monthly payment is $P=\frac{Lc(1+c)^n}{(1+c)^{n}-1}$ , where $P$ is my payment, $L$ is my loan, $c$ is my monthly interest rate, and $n$ is the number of months in my mortgage. This evaluates to be $\frac{300000\times0.005\times1.005^{360}}{(1+0.005)^{360}-1}=\boxed{\$1798.65}$

Present value of mortgage is $300000$ . Let $k$ be the monthly payment. Since payment is made monthly, hence there are $360$ payments in $30$ years. Hence, $PV=300000=kv+kv^2 +...+kv^{360}, v=1/1.005$

Solving the geometric series gives $\boxed{1798.65}$ .

If we borrowed $L$ amount of money with monthly interest rate of $r \text{\%}$ , by $n$ -month, we need to pay back at

$L(1+r \text{\%} )^{n} = L(1+0.01r) ^{n}$ .

If we pay $P$ each month, the total amount of money that we have paid back by $n$ -month is

$P(1+0.01r)^{n-1}+P(1+0.01r)^{n-2}+...+P(1+0.01r)^{1}+P=P\dfrac{(1+0.01r) ^{n}-1}{(1+0.01r)-1)}=P\dfrac{(1+0.01r) ^{n}-1}{0.01r}$

Equating the two equations, we have

$\begin{aligned} P\dfrac{(1+0.01r) ^{n}-1}{0.01r}&=L(1+0.01r) ^{n} \\ P&=L \dfrac{0.01r(1+0.01r)^{n}}{(1+0.01r) ^{n}-1} \end{aligned}$

Putting $r=0.5$ , $n=30 \times 12=360$ , and $L=300000$ , we have $\boxed{P=1798.65}$ .

Is my concept correct?