Question

Find a recurrence relation for the balance owed at the end of months on a loan $5000 of at a rate of 7% if a payment of is made each month. [Hint: Express B(k) in terms of B(k - 1) the monthly interest is (0.07 / 12) B (k - 1).]

Short answer

Given: We can express the B(k) in terms of B (k - 1) the monthly interest is r = 7%

Step-by-step solution

Step 1:

Let represents the balance at the end of months.

The loan is $5000. Thus, the balance at zero months is $5000(when you have not paid anything off yet, nor did the balance increased by the interest).

B (0) = 5000

Each month the balance increases by the interest and decreased by the payment of . Thus, the balance is the balance of the previous month increased by the interest and decreased by the payment of $ 100 .

B (k) = B (k - 1) + I (k) - 100

The interest has a rate of per year, which then corresponds by one twelfth of 7% per month. The interest is the product of the rate per month and the balance of the previous month.

$\begin{array}{r}\mathrm{I}\left(\mathrm{k}\right)=\frac{1}{12}◻\mathrm{\%}\mathrm{B}(\mathrm{k}-1)\\ =\frac{1}{12}◻07\mathrm{B}(\mathrm{k}-1)\\ =\frac{0.07}{12}\mathrm{B}(\mathrm{k}-1)\end{array}$

Step 2:

Replace the found expression of in the equation B(k) = B (k - 1) + I(k) - 100:

$\begin{array}{r}\mathrm{B}\left(\mathrm{k}\right)=\mathrm{B}(\mathrm{k}-1)+\mathrm{I}\left(\mathrm{k}\right)-100=\mathrm{b}(\mathrm{k}-1)+\frac{0.07}{12}\mathrm{B}(\mathrm{k}-1)-100\\ =\left(1+\frac{0.07}{12}\right)\mathrm{B}(\mathrm{k}-1)-100\end{array}$

Hence, we conclude that$\mathrm{B}\left(0\right)=5000\text{and}\mathrm{B}\left(\mathrm{n}\right)=\left(1+\frac{0.07}{12}\right)\mathrm{B}(\mathrm{n}-1)-100$