466
8.2. GEOMETRIC SERIES
As long as P 1 is positive, we still have to keep making payments to pay off the loan.
(a) Recall that the amount of interest we pay each time depends on the principal
that remains. How much interest, in terms of P 1 and r, do we pay in the second
installment?
(b) How much of our second monthly installment goes to pay off the principal?
What is the principal P 2 , or the balance of the loan, that we still have to pay
off after making the second installment of the loan? Write your response in the
form P 2 = ( )P 1 − ( )M, where you fill in the parentheses.
(c) Show that P 2 =
1 +
r
12
2 P −
1 +
1 +
r
12
M.
(d) Let P 3 be the amount of principal that remains after the third installment.
Show that
P 3 =
1 +
r
12
3
P −
1 +
1 +
r
12
+
1 +
r
12
2
M.
(e) If we continue in the manner described in the problems above, then the
remaining principal of our loan after n installments is
P n =
1 +
r
12
n
P −
n−1
k=0
1 +
r
12
k
M.
(8.7)
This is a rather complicated formula and one that is difficult to use. However,
we can simplify the sum if we recognize part of it as a partial sum of a geometric
series. Find a formula for the sum
n−1
k=0
1 +
r
12
k
.
(8.8)
and then a general formula for P n that does not involve a sum.
(f) It is usually more convenient to write our formula for P n in terms of years
rather than months. Show that P(t), the principal remaining after t years, can
be written as
P(t) =
P −
12M
r
1 +
r
12
12t
+
12M
r
.
(8.9)
(g) Now that we have analyzed the general loan situation, we apply formula (8.9) to
an actual loan. Suppose we charge $1,000 on a credit card for holiday expenses.
If our credit card charges 20% interest and we pay only the minimum payment
of $25 each month, how long will it take us to pay off the $1,000 charge? How
much in total will we have paid on this $1,000 charge? How much total interest
will we pay on this loan?
8.2. GEOMETRIC SERIES
As long as P 1 is positive, we still have to keep making payments to pay off the loan.
(a) Recall that the amount of interest we pay each time depends on the principal
that remains. How much interest, in terms of P 1 and r, do we pay in the second
installment?
(b) How much of our second monthly installment goes to pay off the principal?
What is the principal P 2 , or the balance of the loan, that we still have to pay
off after making the second installment of the loan? Write your response in the
form P 2 = ( )P 1 − ( )M, where you fill in the parentheses.
(c) Show that P 2 =
1 +
r
12
2 P −
1 +
1 +
r
12
M.
(d) Let P 3 be the amount of principal that remains after the third installment.
Show that
P 3 =
1 +
r
12
3
P −
1 +
1 +
r
12
+
1 +
r
12
2
M.
(e) If we continue in the manner described in the problems above, then the
remaining principal of our loan after n installments is
P n =
1 +
r
12
n
P −
n−1
k=0
1 +
r
12
k
M.
(8.7)
This is a rather complicated formula and one that is difficult to use. However,
we can simplify the sum if we recognize part of it as a partial sum of a geometric
series. Find a formula for the sum
n−1
k=0
1 +
r
12
k
.
(8.8)
and then a general formula for P n that does not involve a sum.
(f) It is usually more convenient to write our formula for P n in terms of years
rather than months. Show that P(t), the principal remaining after t years, can
be written as
P(t) =
P −
12M
r
1 +
r
12
12t
+
12M
r
.
(8.9)
(g) Now that we have analyzed the general loan situation, we apply formula (8.9) to
an actual loan. Suppose we charge $1,000 on a credit card for holiday expenses.
If our credit card charges 20% interest and we pay only the minimum payment
of $25 each month, how long will it take us to pay off the $1,000 charge? How
much in total will we have paid on this $1,000 charge? How much total interest
will we pay on this loan?