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Section 5.3
143
(b) [BB] The function
f(n)
= 24 is constant, so we
try
Pn
=
k.
This gives
k
=
2k
+
15k
+
24,
so
k
=
2 =
Pn.
In part (a), we saw that the corresponding homogeneous recurrence relation
has solution
qn
=
c1(3
n
)
+
C2( 5)n,
so
an
=
Pn
+
qn
=
c1(3
n
)
+
C2( _5)n

2. The initial
conditions give 1 =
ao
= C1
+
C2 
2 and
1 =
a1
=
3C1

5C2

2, so C1 = 2, C2 = 1 and our
solution is
=
2(3
n
)
+
(_5)n

2.
11. (a) To fit this problem to Theorem 5.3.1, we rewrite the given recurrence as
=
8an1 16an2,
n
?: 2. The characteristic polynomial is
x
2
+
8x
+
16 with repeated characteristic root
4. We
obtain
= C1
(_4)n
+
C2n( 4)n.
Since C1 = 5,
4C1

4C2
= 17, we get C1 = 5, C2 = 37/4.
The solution is
= 5(
_4)n

31
n( _4)n
=
(_4)n 20
4
37n
=
(_4)n1 (37n

20).
(b) Since
= 5 is constant, we
Pn
=
k
and obtain
k
=
8k

16k
+
5, so
k
=
g
=
Pn.
From part (a), we know that the corresponding homogeneous recurrence relation has solution
qn
= C1
+
C2n( _4)n
so
=
Pn
+
qn
= C1
+
C2n( 4)n
+
g. The initial conditions
give 2 =
= C1
+
g
and
=

+
g,
so C1 =
~
and C2 =
lOur solution is
=
~(4)n

~n(
4)n
+
g.
12. (a) [BB] Using Theorem 5.3.1 (or guessing and verifying as in Section 5.2), the solution is
=
4n.
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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