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Section 5.3
145
16. We try
Pn
=
a +bn
and obtain
a+ bn
=
4(a+ b(n
1)) 
 2))
+n,
so
=
4b+n;
that is,
a
=
4,
b
=
1. Thus, we have
=
4
+
n
as a particular solution. As seen in Problem 21, the
homogeneous recurrence
an
=
4an
l

4an2
has solution
qn
=
cl(2
n
)
+
c2n(2n),
so
+
=
4
+
n
+
n
)
+
c2n(2n).
The initial conditions give 4
+
Cl
= 5,5
+
2Cl
+
2C2
=
9, so Cl
=
1,
C2
=
1. Therefore,
=
4
+
n
+
2
n
+
n2n
=
4
+
n
+
2n(1
+
n).
17. [BB] Try
=
a
+
bn
+
cn
2
•
We obtain
a
+
bn
+
2
=
5[a
+
b(n

1)
+
c(n
 1)2]
2[a
+

2)
+

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

b
+
c
+
(b

2c)n
+
2
]


2b
+
4c
+

4c)n
+
2
]
+
2
=
3a

b
 3c
+
(3b

+
(3c
+
3)n
2
.
This gives the system
a
=

b

3c;
b
=
3b

2c;
c
=
3c
+
3
with solution
a
=
3,
b
=
~,
c
=
~.
Hence, a particular solution is
=
3 
~n

~n2.
The homogeneous recurrence
an
=
5an

2an2
has characteristic polynomial
x
2

5x
+
2 with
characteristic roots
5±fY.
Hence,
=
Cl(5f11)n
+
C2(5±f11)n
and
+
=
~n

~n2
+
(5fYf
+
(5±fYf·
The initial conditions give
3
+
+
=
0 and
6
+
Cl(5fY)
+
C2(5±fY)
=
3. This gives
=
~(1

vb),
=
~(1
+
vb)
and, hence,
a
=

2n

2n
2
+
2
(1
__ 1_)
(5_v'I7)n
+
2
+
_1_)
(5±v'I7)n
n
2
2
2
v'I7
2
2
v'I7
2
.
18. We look for a particular solution of the
fonnpn
=
a(5
n
).
Substituting into the recurrence relation, we
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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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