Download Section 13.3 Geometric Sequences

273
Section 13.3 Geometric Sequences
Objective #1: Determining if a sequence is geometric.
If you start with a number a and multiply it a non-zero fixed constant r over
and over again, you would get what is called a geometric sequence.
Definition
Let a and r be real numbers where r ≠ 0. A Geometric Sequence or
Geometric Progression may be defined recursively as
1)
a1 = a and an = ran – 1
The non-zero number r is called the
common ratio. The terms of the geometric ratio will be a, ar, ar2, ar3, …
Determine if the following sequences are geometric:
Ex. 1a
– 3, 6, – 12, 24 ….
Ex. 1b
1, 2, 4, 6, 8, …
Solution:
a)
Since the ratio of each successive terms is – 2 and the first
term is – 3, the sequence is geometric. We can define it as follows:
a1 = – 3
an = – 2an – 1
b)
Since the common ratio between the first two terms is 2, but the
ratio between 5th and 4th terms is 4/3, the sequence is not geometric.
Ex. 2a
{sn} = {5•3n}
Ex. 2b
{pn} = {ln((n + 1)2)}
Solution:
a)
s1 = 5•31 = 15
The ratio between the nth term and the (n – 1)th term is:
sn
sn −1
€
€
=
5•3n
5•3n −1
=3
Thus, the common ratio is 3, so the sequence is geometric. We
can define it as follows:
s1 = 15
sn = 3•sn – 1
Since, ln((n + 1)2) = 2ln(n + 1), then pn= {2ln(n + 1)}
p1 = 2ln(1 + 1) = 2ln(2)
The ratio between the nth term and the (n – 1)th term is:
b)
sn
sn −1
=
2ln(n+1)
2ln(n−1+1)
=
ln(n+1)
ln(n)
which is not constant, so the
sequence is not geometric.
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€
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Objective #2
Deriving a formula for geometric sequences.
To derive the formula for geometric sequence, let’s consider the first few
terms of the sequence.
a1 = a
a2 = a1 r
a3 = a2r = a1rr = a1r2 = a1r(3 – 1)
a4 = a3r = a1r2r = a1r3 = a1r(4 – 1)
a5 = a4r = a1r3r = a1r4 = a1r(5 – 1)
⋮
⋮
an = an – 1 r = …
= a1r(n – 1)
Theorem
Let a and r be real numbers, r ≠ 0. Then the nth term of a geometric
sequence is given by the formula: 2) an = a1r(n – 1).
Find the indicated term in each Geometric sequence:
Ex. 3
7th term of 4, 2, 1, ½, ….
Solution:
The first term is a1 = 4 and r = ½. By formula #2, the nth term is
an = 4(½)(n – 1). Thus, the 7th term is
1
1
a7 = 4(½)(7 – 1) = 4(½)6 = 4( ) =
= 0.0625
64
16
Find the following:
Ex. 4
The 4th term is 7 and the 7th term is – 23,625
€ term€and the common ratio.
a)
Find the first
b)
Give a recursive formula for the sequence.
c)
What is the nth term of the sequence?
Solution:
a)
From formula #2, an = ar(n – 1). Since a4 = 7 and a6 = – 23,625,
then 1)
a4 = ar3 = 7
2)
a7 = ar6 = – 23,625
Dividing the second equation by the first yields:
r3 = – 3375
(Take the "cube root" of both sides)
r = – 15
Now, replace r = – 15 in equation #1 to solve for a1:
7
a(– 15)3 = 7
⇒
– 3375a = 7
⇒
a=–
Thus, the first term is –
7
3375
3375
and the common ratio is – 15.
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€
275
b)
Plugging in a1 = –
a1 = –
c)
7
3375
and r = – 15 into formula #1 yields:
an = – 15an – 1
Plugging €
in a1 = –
€ an = –
7
3375
7
(–
3375
7
3375
and r = – 15 into formula #2 yields:
15)(n – 1) = –
(– 15)(n – 1) =
3
−(−15)
(n – 1 – 3)
= 7(– 15)
= 7(– 15)
€
Thus, an = 7(– 15)(n – 4).
€
Objective #3
7
7
3
(−15)
(– 15)(n – 1)
(n – 4)
€ of a geometric sequence.
€
Find the sum
Theorem: Sum of the First n Terms of a Geometric Sequence
Let {an} be a geometric sequence with the first term a1 and common ratio r
where r ≠ 0 and r ≠ 1. The sum Sn of the first n terms of {an} is given by:
3)
Sn = a1
(
1 − rn
1− r
),
r ≠ 0, 1
Proof:
By definition,
Sn = a1 + a2 + a3 + … + an (use formula #2 for each term)
4) €Sn = a1 + a1r + a1r2 + … + a1r(n – 1)
(multiply both sides by – r)
2
(n – 1)
n
5) – rSn = – a1r – a1r – … – a1r
– a1r (add equation #4 to #5)
n
Sn – rSn = a1 – a1r
(factor out Sn and a1)
n
(1 – r)Sn = a1•(1 – r )
(divide both sides by (1 – r)
Sn = a1
(
1 − rn
1− r
)
r ≠ 0, 1
Find the sum of the first n terms of the following sequence:
Ex. 5
{an} = {5•3(n – 1)}
€
Solution:
The first term is a1 = 5•30 = 5 and the common ratio is r = 3.
Using formula #3, we get:
Sn = 5
n
n
( 11−− 33 ) = 5( 1−− 23 ) = – 2.5(1 – 3 ) = 2.5•3
n
n
– 2.5
– 1 – 2 – 4 – 8 – … – 214
Solution:
€ The first term
€ is a = – 1 and the common ratio is r = 2
1
Ex. 6
276
Using formula #3, we get:
Sn = – 1
(
1 − 2n
1− 2
)
=–1
(
1 − 2n
−1
)=1–2
n
Since an = – 214 = – 1(2)(n – 1) by formula #2, then n – 1 = 14 or n = 15.
Thus, S15 = 1 – 215 = 1 – 32,768 = – 32,767
Ex. 7
€
3+
€12
+
+…+3
6
5
25
12
( 25 )
Solution:
The first term is a1 = 3 and the common ratio is r =
€ €formula #3, we
€ get:
Using
Sn = 3
(
2
5
2
5
1 − ( )n
1−
Since an = 3
) (
=3
12
( 25 )
2
5
3
5
1 − ( )n
)
= 3•
5
3
[1 –
2 n
]
€5
( )
2
5
=5–5
2 n
5
( )
n–1
( 25 ) by formula #2, then n – 1 = 12 or
2 13
€
= 5 – 5(€ ) = €5 – 0.000033554432
5
=3
€ n = 13. Thus,
€ S13
= 4.999966445568
€
€
Objective #4:
Determining if a geometric series converges or diverges.
€
In the last example, suppose we did not stop with just adding the first 13
terms. What if we added the first 20 terms or 100 terms or 1000 terms? We
might guess that we would get an answer very close to five. If we let
number of terms in the sequence increase without bound, would the sum
be five? If so, when would the sum of all terms of geometric series
approach a number (converges) or increase without bound (diverges) when
the number of terms increases without bound? To answer, we need to
define what we mean by an infinite series.
Definition
Let a1 and r be real numbers where r ≠ 0. An Infinite Geometric Series is
defined as:
∞
∑ a1r k−1 = a
1
+ a1r + a1r2 + … + a1r(n – 1) + …
k=1
where a1 is the first term and r is called the common ratio.
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Definition
Let Sn be the sum of the n terms of a geometric series. As n increases
∞
without bound, if Sn approaches a finite number L, then
∑ a1r k−1 = L and
k=1
we say that the series converges. If a series does not converge, then it
diverges. Recall that Sn = a1
fractions:
a1
(
1 − rn
1− r
)
=
a1
1− r
(
–
1 − rn
1− r
a1r n
.
1− r
). We can rewrite this as two separate
€
The first term is fixed so its value will
not change as n increases without bound. For the second term, if |r| > 1,
€ increase without bound as n is increases without
then that term will also
bound. Thus, the series will diverge in that case. If 0 < |r| < 1, then the
€ get closer
€ term will
€
second
to zero as n increases without bound, In this
∞
instance, the series will converge and
∑ a1r k−1 = 1a− r .
1
k=1
Theorem Convergence of an infinite geometric series
€ ∞
€
If 0 < |r| < 1, then the infinite geometric series a1r k−1 converges to
∑
k=1
a1
.
1− r
Determine if the following geometric series converges or diverges. If
€
it converges, find its sum.
€
6
12
2 n–1
Ex. 8
3+
+
+…+3
+…
5
(5)
25
Solution:
We speculated that this might converge and that the sum might
approach
without bound earlier. Now, since r =
€ € 5 as n increases
€
2
5
,
then by the convergence of an infinite geometric series theorem, the
2
series does converge. Since a1 = 3 and r = , then the sum will
5
equal:
€
∞
∑ 3( 25 )k−1 = 1a− r = 1 −3
1
k=1
€
€
€
2
5
€
=3÷
€
3
5
= 5.
278
∞
∑ 0.6( 76 )k−1
Ex. 9
k=1
Solution:
Since r =
7
6
> 1, then the series diverges.
€
Show the following:
1
Ex. 10
Show that the repeating decimal 0.3 = 0.333… = .
3
€
Solution:
Since 0.333… = 0.3 + 0.03 + 0.003 + … + 0.3(0.1)n – 1 + …,
then we can represent this as the infinite geometric series:
€
∞
∑ 0.3(0.1)k−1
where a1 = 0.3 and r = 0.1
k=1
Since r = 0.1 < 1, the series converges and the sum is equal to:
∞
∑ 0.3(0.1)k−1 = 1a− r = 1 −0.30.1 =
1
€
k=1
0.3
0.9
Thus, the repeating decimal 0.333… =
Objective #5:
€
=
3
9
=
1
3
1
.
3
€ €
€
€
Solving
annuity €
problems.
If we invested a fixed amount in an account
paying compound interest, we
€
used a formula previously to calculate how much will be in the account after
a certain period of time. Many times however, people do not invest a fixed
amount, but make periodic deposits into an account. An annuity is a
sequence of equal periodic deposits. If the deposits are made at the same
time that interest is calculated, then the annuity is called an ordinary
annuity. With such an annuity, if interest is compounded monthly, then the
deposits are made monthly at the same time. If interest is compounded
weekly, then the deposits are made weekly at the same time and so forth.
The amount of the annuity is the sum of all the deposits made plus all the
interest earned.
When working with annuity, we will need to calculate the interest rate
earned per pay period. Thus, if the account pays 6% interest compounded
quarterly, then the interest rate per pay period is:
0.06
i=
= 0.015.
4
€
279
The number n of pay periods is equal to the number of times interest is
calculated per year times the number of years. Thus, if interest is
compounded quarterly for five years, the number of payment periods will be
n = 4•5 = 20
Let P be the amount of money deposited at each payment period for n
payment periods in an account that earns i percent interest per payment
period. When the last deposit of P is made at the nth payment, the amount
that was deposited at the first payment period would have earned interest
for n – 1 payment periods. The amount of money that first deposit becomes
at the end of the annuity is P(1 + i)n – 1. The amount that was deposited at
the second payment period would have earned interest for n – 2 payment
periods. The amount of money that first deposit becomes at the end of the
annuity is P(1 + i)n – 2. We can continue this pattern for all the payments.
The amount A of the annuity is just the sum of all the amounts generated
from the deposits from each payment period:
A = P(1 + i)n – 1 + P(1 + i)n – 2 + … + P(1 + i)1 + P
(reorder)
= P + P(1 + i)1 + … + P(1 + i)n – 2 + P(1 + i)n – 1
This is the sum of a geometric series with the first term of a1 = P, the
common ratio of r = (1 + i). We can apply formula #3 to get:
n
A=
∑
P(1 + i)k−1 = P•
(
k=1
1 − (1 + i)n
1 − (1 + i)
)
= P•
(
1 − (1 + i)n
1− 1− i
)
=P
(
(1 + i)n − 1
i
)
Theorem Amount of an Annuity
Let P be the amount deposited at the end of each payment period for an
€
€ pays i percent€interest per payment
annuity. If the annuity
period, then the
€
amount A of the annuity after n deposits is:
A=P
n
( (1 + i)i − 1 )
Solve the following:
Ex. 11
Leroy contributes $250 at the end of each quarter to his taxsheltered annuity. What will be the value of his annuity after the 80 th
€ (20 years) if interest is 8% compounded quarterly?
deposit
Solution:
Plugging P = $250, i = 0.08/4 = 0.02, and n = 80 into the formula, we
get:
A=P
n
80
( (1 + i)i − 1 ) = (250)( (1 + 0.02)
0.02
−1
His annuity will be worth $48,442.99.
€
€
≈ 48,442.99
) = 250( 3.8754...
0.02 )
€
280
Ex. 13
Juanita decides instead of buying a new car every five years
and being stuck making car payments of $400 per month, she
decides to invest the $400 per month in a mutual fund that
historically returns 9% growth compounded monthly. If she does this
for forty years, how much will she have in her mutual fund at the end
of 40 years?
Solution:
Plugging P = $400, i = 0.09/12 = 0.0075, and n = 12•40 = 480 into the
formula, we get:
A=P
(
(1 + i)n − 1
i
)
= (400)
(
(1 + 0.0075) 480 − 1
0.0075
) = 400( 35.1099...
0.0075 )
≈ 1872528.11
Her mutual fund will be worth $1,872,528.11.
€
€
€