Loan and investment problems offer great applications of recursive sequences.
PAYING BACK A LOAN
SAVING FOR THE FUTURE
This section requires calculations that cannot reasonably be done by hand.
Your teacher can show you how to work with recursive sequences on your calculator.
Suppose you are borrowing $\,\$22{,}000\,$.
Interest is being charged at an annual rate of $\,5\%\,$.
You plan to pay back $\,\$250\,$ each month; this payment goes to both interest and principal.
(a) Find the interest owed in the first month.
(b) Write a recursive formula where $\,u_n\,$
gives the amount owed after $\,n\,$ months.
(c) Then, find the amount owed after one year of payback.
(d) Find the total principal paid in the first year.
(e) Find the total interest paid in the first year.
(f) Putting it all together—the brief solution.
(a)
The annual interest rate is $\,5\% = 0.05\,$.
The monthly interest rate is $\,\frac{0.05}{12}\,$.
The interest owed in the first month is $\,(\$22{,}000)(\frac{0.05}{12}) = \$91.67\,$.
Notice that your monthly payment must be at least $\,\$91.67\,$ to cover the interest.
If you were to pay exactly $\,\$91.67\,$ the first month, then your debt remains $\,\$22{,}000\,$.
If you were to pay $\,\$91.67\,$ each month for the next twenty years, you'd still owe $\,\$22{,}000\,$ at the end of those twenty years.
Obviously, you want to make sure you pay more than this each month, so the amount of debt decreases.
You've decided to pay $\,\$250\,$ each month, so with this first payment you reduce your debt by $\,\$250 - \$91.67 = \$158.33\,$.
That means you now owe $\,\$22{,}000 - \$158.33 = \$21{,}841.67\,$, which will accrue a smidgeon less interest for the next month.
Look at the pattern for the amount you owe at the end of the first month.
Dollar signs and commas are suppressed, to make it a bit easier on the eyes:
$$
\overset{\text{amt owed at beginning}}{\overbrace{\strut 22000}} +
\overset{\text{interest accrued for the month}}{\overbrace{\strut (22000)(\frac{0.05}{12})}} -
\overset{\text{monthly payment}}{\overbrace{\strut \ \ 250\ \ }}
=
\overset{\text{amount owed after one month}}{\overbrace{\strut 21841.67}}
$$
Now we'll introduce the notation.
Let $\,u_0\,$ denote the amount owed at time zero; that is, the initial debt.
Let $\,u_1\,$ denote the amount owed after $\,1\,$ month (i.e., after $\,1\,$ payment).
Rewriting the formula above using this notation gives:
$$
\overset{\text{amt owed at beginning}}{\overbrace{\strut \ \ u_0\ \ }} +
\overset{\text{interest accrued for the month}}{\overbrace{\strut (u_0)(\frac{0.05}{12})}} -
\overset{\text{monthly payment}}{\overbrace{\strut \ \ 250\ \ }}
=
\overset{\text{amount owed after one month}}{\overbrace{\strut\ \ u_1\ \ }}
$$
Rewrite the equation from right-to-left and factor, to get:
$$
u_1 = (1 + \frac{0.05}{12})u_0 - 250
$$
(b)
The pattern above gets applied month after month after month.
Take the prior amount owed: $\,u_{n-1}\,$
Add in the interest accrued on this amount: $\,u_{n-1}(\frac{0.05}{12})\,$
Subtract off your payment: $\,250\,$
Thus, the amount owed after $\,n\,$ months is:
$$\,u_n = (1 + \frac{0.05}{12})u_{n-1} - 250\,,\ \ \text{ for } n\ge 1$$
(c)
Note: For this part of the problem, you need a calculator that does recursion.
The amount owed after one year (twelve months) is $\,u_{12}\,$.
From the calculator:
$\,u_{12} = 20055.85\,$
Thus, you owe $\,\$20{,}055.85\,$ after one year.
(d)
The total principal paid in the first year is: $\,\$22{,}000 - \$20{,}055.85 = \$1944.15$
(e)
How much money did you actually send to the bank in the first year?
Twelve payments of \$250 each: $\,(12)(\$250) = \$3000\,$
Of this, you know from part (d) that $\,\$1944.15\,$ went to principal—that is, reduced your debt.
The remainder is interest—your payment for the privilege of using the bank's money for the year.
Thus, the total interest paid in the first year is: $\,(12)(\$250) - \$1944.15 = \$1055.85$
(f)
Here's the brief solution:
(a) The interest owed in the first month is: $\,(\$22{,}000)(\frac{0.05}{12}) = \$91.67\,$
(b) The recursive formula is:
$u_0 = 22000$
$\,u_n = (1 + \frac{0.05}{12})u_{n-1} - 250\,,\ \ \text{ for } n\ge 1$
(c) From the calculator: $\,u_{12} = \$20{,}055.85\,$
(d) The total principal paid in the first year is: $\,\$22{,}000 - \$20{,}055.85 = \$1944.15$
(e) The total interest paid in the first year is: $\,(12)(\$250) - \$1944.15 = \$1055.85$
You are saving for the future.
Your initial deposit is $\,\$4100\,$.
Interest is being earned at an annual rate of $\,5\%\,$, compounded monthly.
You will contribute an additional $\,\$120\,$ each month.
(a) Find the interest earned in the first month.
(b) Write a recursive formula where $\,u_n\,$ gives the amount saved (principal plus interest) after $\,n\,$ months.
(c) Then, find the amount saved (principal plus interest) after $\,7\,$ years.
(d) Find the total amount of money you contributed (principal only) during these $\,7\,$ years.
(e) Find the total interest earned during these $\,7\,$ years.
SOLUTION:
(a) The interest earned in the first month is: $(\$4{,}100)(\frac{0.05}{12}) = \$17.08$
(b) The recursive formula is:
$u_0 = 4100$
$u_n = (1 + \frac{0.05}{12})u_{n-1} + 120\,$, for $\,n \ge 1$
(c) Note that $\,7\,$ years is $\,7(12) = 84\,$ months.
From the calculator: $\,u_{84} = 17853.39$
Thus, you have saved (principal plus interest) $\,\$17{,}853.39\,$ after $\,7\,$ years.
(d) The total amount of money you contributed (principal only) during these $\,7\,$
years is: $\,\$4100 + 7(12)(\$120) = \$14{,}180.00$
(e) The total interest earned during these $\,7\,$ years is:
$\,\$17{,}853.39 - \$14{,}180.00 = \$3,673.39$