Math 368-1
Second Exam SOLUTIONS 1, 2, 4-7
Summer I, 2008
1. Suppose n is a positive integer and r is a real number.
(a) If r ≠ 1, write down a formula for 1 + r + ... + rn-1 . (Proof not necessary.)
SOLUTION. 1 + r + ... + rn-1 =
rn −1
.
r −1
(b) What is the sum if r = 1? Explain!
SOLUTION. When r = 1, 1 + r + ... + rn-1 = 1 + 1 + ... + 1n-1 = 1 + 1 + ... + 1 (n times) = n.
2. (a) In an ordinary annuity, you make a periodic payment P with periodic interest rate i.
Write down the basic recurrence relation which relates the amount in the account, Pn+1 , after
n+1 periods to the amount in the account, Pn , after n periods.
SOLUTION. Since a payment P is made at the END of each period, the amount in the account at the
end of period n + 1 is
Pn+1 = (1+i)Pn + P (for n ≥ 1).
(b) In an amortized loan, you borrow an amount Po and make a periodic payment M with periodic
interest rate i. Write down the basic recurrence relation which relates the balance on the loan,
Pn+1 , after n+1 periods to the balance on the loan, Pn , after n periods.
SOLUTION. Since a payment M is made at the END of each period, the balance owed at the end of
period N + 1 is
Pn+1 = (1+i)Pn - M (for n ≥ 0).
3. Prove ONE of the following formulas by induction, using the recurrence relation from the preceding
problem:
(1 + i) n − 1
i
SOLUTION. Not given here.
Ordinary annuity:
Pn = P
Amortized loan:
Pn = Po(1+i)n - M
(1 + i) n − 1
i
4. Refer to the ordinary annuity situation. What is the present value of the balance in the account at
the end of n periods?
SOLUTION. present value = amount at the end of n periods discounted to the present
= Pn(1+i)-n = P
(1 + i) n − 1
i (1+ i) n
= P
1− (1+ i )−n
i
5. Solve the following problem completely without using a calculator, using ONLY the fact that
(1.05)10 = 1.62889462678.
Chris and Robin get married today. Chris immediately starts saving $1000 per year in an account
which pays 3.25% compounded annually. Chris makes 15 such payments, and 15 years from now,
the money is put into an account paying 5% per year, compounded annually, for 10 years.
Ten years from now, Robin starts a similar program, putting an amount A into an account paying
3.25% per year, and Robin also makes 15 deposits.
When Chris and Robin get divorced 25 years from now, it turns out that they each have exactly the
same amount of money in their savings accounts. What was the amount A that Robin was
depositing?
SOLUTION. Let i =0 .0325
Amount in Chris’ account after 15 years:
$1000[(1+i)15 + (1+i)14 + … + (1+i)1 ]
Amount in Chris’ account 10 years later, after 25 years total:
$1000[(1+i)15 + (1+i)14 + … + (1+i)1 ] (1 + .05)10.
Amount in Robin’s account after 15 years, 25 years from now:
A[(1+i)15 + (1+i)14 + … + (1+i)1 ]
Since Chris and Robin have the same amount 25 years from now,
$1000[(1+i)15 + (1+i)14 + … + (1+i)1 ] (1 + .05)10
= A[(1+i)15 + (1+i)14 + … + (1+i)1 ]
so, since all the terms with i in them cancel out,
A = $1000(1.05)10 = $1000(1.62889) = $1628.89.
6. For reasons which we will not go into here, your monthly mortgage payment decreases by an amount
$40 per month, for the next 15 years. You decide to use the savings to get a new loan, borrowing a
certain amount at 6% compounded monthly, making monthly payments of $40 on the new loan for
the next 15 years. How much can you afford to borrow?
SOLUTION. We use the formula from Problem 4 for the present value of a 15 year annuity (180
months) at 6%, with $40 monthly payments:
present value = P
1− (1+ i )−n
i
= 40
1− (1+ .06/12)−180
.06 /12
= $4740.14.
7. The table below shows the first row of a monthly amortization schedule.
(a) How much money was borrowed?
(b) What is the annual percentage rate of the mortgage?
Month
Monthly
Payment
Interest
Paid
1
$376.24
$110.86
Principal
Repaid
$265.40
Outstanding
Principal
$18734.60
SOLUTION. (a) amount borrowed = first month’s principal repaid + first month’s ending balance
= $19000.
(b) Let r = annual interest rate. Then
interest paid in first month = (r/12)(original balance),
so r = 12(110.86/19000) ≈ 0.07 = 7%.