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Section 5.4
(b) We have
as desired.
2x[c(n
1)x
n

1
J
x
2
[c(n

2)x
n

2
J
2c(n

l)xn

c(n

2)xn
=
cnx
n
=
an
149
(c) By part (b),
C2nXn
is one solution and by part (a) of Exercise 25
CIX
n
is another. Hence, by part (b)
of Exercise 25, we conclude that
CIX
n
+
c2nx
n
is also a solution. Finally, two initial conditions
will detennine
Cl and C2 because they detennine two linear equations in the two unknowns Cl, C2
which are guaranteed to have a unique solution.
27. (a) We have
Pn
=
rPnl
+
SPn2
+
f(n)
for
n
:2:
2 and also
qn
=
rqnl
+
sqn2
for
n
:2:
2. Hence,
rPnl
+
SPn2
+
f(n)
+
rqnl
+
sqn2
r(Pnl
+
qnl)
+
S(Pn2
+
qn2)
+
f(n)
which says that
Pn
+
qn
satisfies the given relation for
n
:2:
2.
(b) Part (a) shows that
Pn
+
qn
satisfies the given recurrence relation where
qn
=
CIXl'
+
C2X~
or
qn
=
CIX
n
+
C2nXn
according as the polynomial
x
2

rx

s
has two distinct roots or just one.
Two initial conditions provide two equations which can be solved for
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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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