A bank has a loan scheme with an annual interest rate of $0<r<1$ ($100r$ is the annual percentage rate). The bank charges interest every month. You take a loan of $L$ pounds and wish to pay it back in exactly $m$ months by making monthly payments of $p$ pounds each. Find $p$.
1. Evaluate the sum $a+ar+ar^2+\cdots+ar^n,$ with $r \neq 0$ and $n>0.$
2. Given $a_0= 5$ and $a_n= 3a_{n-1}+4,$ find $a_4.$

Find an expression for the amount remaining to be paid after month $i.$

… in terms of the amount remaining to be paid in month $i-1.$

You might first want to find the cumulative monthly interest rate. Hint: it’s not $r/12.$

… knowing that the cumulative monthly interest rate is the interest rate that gives $r$ when applied successively $12$ times.

If $x_i$ denotes the amount remaining to be paid after month $i,$ you should now have a recursive formula for $x_i.$ Try to find the explicit formula.

You should get an expression that contains $x_0.$ What is the value of $x_0?$

Let $x_i$ be the amount remaining to be paid after month $i$. Thus, $x_0=L.$ We have $x_i=x_{i-1}(1+\mu)-p,$ where $\mu$ is the cumulative monthly interest rate (we’ll determine its value later). To simplify our working, let $c=1+\mu.$ Unwinding the recursion, we get: \begin{aligned} x_i &= x_{i-1}c-p \\ &= (x_{i-2}c-p)c - p \\ &= \left(\left((x_{i-3}c-p)c - p\right) \right)c-p \\ &= \ldots \\ &= x_0c^i-p\sum_{k=0}^{i-1}c^k \\ &= Lc^i-p\frac{c^i-1}{c-1}. \end{aligned}

We want $x_m=0,$ which gives $p=L\;c^m\;\frac{c-1}{c^m-1}.$

Now we need to express $\mu$ in terms of $r.$ The cumulative monthly interest rate $\mu$ is the interest rate that gives $r$ when applied successively (i.e. recursively) $12$ times. This is not equal to $r/12\;^{(*)}.$ An initial amount $z_0$ becomes $z_0(1+\mu)$ after a month, and $z_0(1+\mu)^{12}$ after $12$ months, which must also be equal to $z_0(1+r).$ This is in fact another recursion: $z_i=z_{i-1}(1+\mu),$ where we have $z_{12}=z_0(1+\mu)^{12}=z_0(1+r).$ Hence, $\mu=(1+r)^{1/12}-1.$

The monthly payment sought is then $p=L\;(1+r)^{m/12}\;\frac{(1+r)^{1/12}-1}{(1+r)^{m/12}-1}.$

(*) $\mu$ can in fact be approximated by $r/12$ for $r\ll1$ though Taylor Series. This is a safe assumption for savings accounts, but not for loans (hmm, what does that tell you about banks?).