$0<r<1$
($100r$
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 $p$
.
-
- Evaluate the sum
$a+ar+ar^2+\cdots+ar^n,$
with$r \neq 0$
and$n>0.$
- Given
$a_0= 5$
and$a_n= 3a_{n-1}+4,$
find$a_4.$
- Evaluate the sum
-
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?).