f.
r
2
+ 6
r
+ 9 = (
r
+ 3)
2
.
So
a
n
=
c
1
(

3)
n
+
c
2
n
(

3)
n
.
We must have
c
1
= 3 and

3
c
1

3
c
2
=

3 so
c
2
=

2.
a
n
= (3

2
n
)(

3)
n
.
g.
r
2
+ 4
r

5 = (
r

1)(
r
+ 5)
.
So
a
n
=
c
1
+
c
2
(

5)
n
.
We must have
c
1
+
c
2
= 2 and
c
1

5
c
2
= 8 so
c
1
= 3
and
c
2
=

1.
a
n
= 3

(

5)
n
.
8.
a.
L
n
=
L
n

1
+
L
n

2
2
.
b.
The characteristic equation of the recurrence is
r
2

1
2
r

1
2
=
1
2
(2
r
2

r

1) =
1
2
(2
r
+ 1)(
r

1). This has
roots

1
/
2 and 1.
So
L
n
=
↵
1
(

1
/
2)
n
+
↵
2
(1
n
) =
↵
1
(

1
/
2)
n
+
↵
2
. Using 100000 =
L
1
=

↵
1
/
2 +
↵
2
and
300000 =
L
2
=
↵
1
/
4 +
↵
2
we get
↵
1
=
800000
3
and
↵
2
=
700000
3
.
Therefore
L
n
=
800000
3
(

1
2
)
n
+
700000
3
.
12.
The characteristic equation of
a
n
= 2
a
n

1
+
a
n

2

2
a
n

3
is
r
3

2
r
2

r
+ 2 = (
r

1)(
r

2)(
r
+ 1), which has
roots 1
,
2
,

1. Therefore we can write
a
n
=
↵
1
·
1
n
+
↵
2
·
2
n
+
↵
3
·
(

1)
n
.
Using the initial conditions, we have
3 =
a
0
=
↵
1
+
↵
2
+
↵
3
6 =
a
1
=
↵
1
+ 2
↵
2

↵
3
0 =
a
2
=
↵
1
+ 4
↵
2
+
↵
3
.
We solve this, to get
↵
1
= 6
,
↵
2
=

1
,
↵
3
=

2. Therefore
a
n
= 6

2
n

2(

1)
n
.
§
8.4
30.
Let
G
(
x
) =
P
k
≥
0
a
k
x
k
.
a.
Then
P
k
2
a
k
x
k
= 2
G
(
x
).
b.
a
0
x
+
a
1
x
2
+
a
2
x
3
+
· · ·
= 2
G
(
x
).
c.
a
2
x
4
+
a
3
x
5
+
a
4
x
6
+
· · ·
=
x
2
(
a
2
x
2
+
a
3
x
3
+
a
4
x
4
+
. . .
) =
x
2
(
G
(
x
)

a
0

a
1
x
)
.
d.
a
2
+
a
3
x
+
a
4
x
2
+
· · ·
=
1
x
2
(
a
2
x
2
+
a
3
x
3
+
a
4
x
4
+
. . .
) =
1
x
2
(
G
(
x
)

a
0

a
1
x
)
.
e.
If
G
(
x
) = (
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
. . .
) then
G
0
(
x
) = (
a
1
+ 2
a
2
x
+ 3
a
3
x
2
+
. . .
)
.
f.
G
(
x
)
2
= (
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
. . .
)
2
= (
a
2
0
+ 2
a
0
a
1
x
+ (2
a
0
a
2
+
a
2
1
)
x
2
+ (2
a
0
a
3
+ 2
a
1
a
2
)
x
3
+
. . .
)
.
32.
Use generating functions to solve the recurrence relation
a
k
= 7
a
k

1
with the initial condition
a
0
= 5.
Let
G
(
x
) =
P
k
≥
0
a
k
x
k
. From the recurrence, we have
a
k
x
k
= 7
a
k

1
x
k
.
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 Math, Zagreb, AirTrain Newark, P1 P2 P3, P1 P2, ak xk, P3 P4