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Find a recurrence relation for the amount of money outstanding on a \$40,000 mortgage after n years. The interest rate on the mortgage is 10% and the yearly payment is \$2,000( the yearly payment is paid at the end of each year after the interest has been computed).

I need someone help with this question? I think the answer is: $a_n = a_{n-1}-(.1)a_{n-1}-2000$.

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    $\begingroup$ Close. Until the last payment, which will be smaller, we have $a_n=(1.1)a_{n-1}-2000$. Similar to yours, but you had a minus sign. $\endgroup$
    – André Nicolas
    Apr 25, 2015 at 20:01
  • $\begingroup$ @AndréNicolas: how can I find what the last payment will be? $\endgroup$
    – tonicPkmn
    Apr 25, 2015 at 21:30
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    $\begingroup$ Either compute the $a_k$, using the recurrence, until $a_k \lt \frac{2000}{1.01}$, ( nuisance, since the mortgage continues for a long time) and then it will be easy, or find a general formula for $a_k$ (forgetting about the fact the mortgage ends). If you find such a formula, you will be able to find the largest $k$ such that $a_k\lt \frac{2000}{101}$ by using logarithms. $\endgroup$
    – André Nicolas
    Apr 25, 2015 at 21:41
  • $\begingroup$ As a quick check, $2,000/0.1=20,000<40,000$. This means that even if the annual repayment of $2,000$ is made forever, the present value of the repayment stream will be less than the value of the mortgage today. Hence the mortgage will never be fully repaid. In fact the amount owed balloons over time. The minimum repayment required for the mortgage to fully discharged is $4,000$ which has to be paid in perpetuity. If the repayment is less than $4,000$, then the mortgage can be fully repaid within a finite time. $\endgroup$
    – Hypergeometricx
    Apr 26, 2015 at 15:32

1 Answer 1

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Assume WLOG that monetary values are $'000.

Let
$r=$ interest rate $=0.1$,
$R=$annual repayment $=2$,
$a_n=$ outstanding mortgage balance at year $n$ after repayment at year $n$,
$P=a_0=$ opening mortgage balance $=40$.

The recurrence relation is $$a_n=a_{n-1}(1+r)-R$$ Rearranging gives $$\begin{align} a_n-\frac Rr&=(1+r)\left(a_{n-1}-\frac Rr\right)\\ &=(1+r)^2\left(a_{n-2}-\frac Rr\right)\\ &\qquad\vdots\\ &=(1+r)^n\left(a_{0}-\frac Rr\right)\\ a_n&=(1+r)^na_0-\frac Rr[(1+r)^n-1]\\ &=\left(P-\frac Rr\right)(1+r)^n+\frac Rr\\ &=20(1.1^n+1)\qquad\blacksquare \end{align}$$ which is the closed form solution for $a_n$.

Note that $a_n$ increases as $n$ increases, hence the outstanding mortgage balance will balloon over time and the mortgage will never be fully repaid.

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