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142
Solutions to Exercises
Exercises 5.3
1. [BB] The characteristic polynomial is
x
2

X

6; the characteristic roots are 2,3. We obtain
an
=
cI(2)n
+
c2(3
n
).
Since CI + C2
=
1, 2CI +
3C2
=
3, we find CI
=
0, C2
=
1. The solution is
an
=
3
n
.
2. The characteristic polynomial is
x
2
+
6x

7; the characteristic roots are
7, 1. We obtain
an
=
cI(7)n
+
c2(ln).
Since CI + C2
=
32, 7CI + C2
= 17,
we find CI
= 49/8,
C2
=
207/8. The
solution is
3. [BB] The characteristic polynomial here,
x
2
6x+9,
has the repeated root
x
=
3 and so Theorem 5.3.1
tells us that the solution is
an
=
cI(3
n
)
+
c2n(3
n
).
From the given values of
ao
and
aI,
we obtain
cI(30) + c2(0)(30)
=
5 (so CI
=
5) and cI(3
1
) + c2(1)(3
1
)
=
3; that is, 3CI +
3C2
=
3. SO
CI + C2
=
1 and C2
=
1 
CI
=
6. The solution is
an
=
5(3
n
)
+
6n(3
n
).
4. To fit this problem to the notation of Theorem 5.3.1, we rewrite the given recurrence in the equivalent
form
an
=
7an
1
1Oan2, n
;::: 3. The characteristic polynomial is
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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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