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
Cl,
C2.
Exercises 5.4
1.
(a) [BB] Since
(2

3x)2
=
4 
12x
+
9x
2
the associated sequence is
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 Summer '10
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 Graph Theory, Recurrence relation, initial conditions, Generating function, associated sequence

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