144
Solutions to Exercises
14. (a) The characteristic polynomial is
x
2
+
6x
+
9 with repeated characteristic root
3. Hence,
an
=
C1
(_3)n
+
C2n( 3)n.
SO
C1
=
1,
3C1

3C2
=
4;
that is,
C1
=
1, C2
=
1/3. Thus,
an
=
(_3)n
+
In( 3)n
=
(3)n(1
+
n/3)
=
(n
+
3)(
_3)n1.
(b) Try
Pn
=
a
+
bn
+
cn
2
.
We obtain
a
+
bn
+
cn
2
=
6[a
+
b(n

1)
+
c(n

1)2]
9[a
+
b(n

2)
+
c(n

2)2]
+
n
2
+
3n
=
6[a

b
+
c
+
(b

2c)n
+
cn
2
]

9[a

2b
+
4c
+
(b

4c)n
+
cn
2
]
+
n
2
+
3n
=
15a
+
24b

42c
+
(15b
+
48c
+
3)n
+
(15c
+
l)n
2
.
Solving, we obtain
a
=
1
5
2
1
8'
b
=
~,
c
=
11
6
,
so
Pn
=
1
5
2
1
8
+
~n
+
1~n2.
From part (a),
qn
=
C1
(_3)n
+
C2n( _3)n,
so
Pn
+
qn
=
t2
1
8
+
~n
+
116n2
+
c1(3)n
+
c2n(3)n.
Using the initial conditions, we obtain
51
_
179
d
51
3
1
3
3
_
21
128
+
c1 
128
an
128
+
8"
+
16 
C1 
C2 

128 '
so
C1
=
1, C2
=
~.
Thus,
an
=
1
5
2
1
8
+
~n
+
l6 n
2
+
(3)n(1

~n).
(c)
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 Summer '10
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 Graph Theory, Characteristic polynomial, Recurrence relation, pn

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