1
Question 1 (10 points)
(a)
(2 points)
Give an example of a linear secondorder homogeneous re
currence with constant coefficients, different from any other such re
currences appearing in this exam.
Answer
: Many possible answers, one that keeps things simple is
a
n
+2
=
4
a
n
+1

3
a
n
.
(b)
(2 points)
If
a
0
, a
1
, a
2
, . . .
is a sequence satisfying this recurrence and
a
0
=
a
1
= 1, what is the value of
a
5
?
Answer
:
a
2
= 4
·
1

3
·
1 = 1, and similarly
a
3
=
a
4
=
a
5
= 1.
(c)
(3 points)
What is the general solution to the recurrence you provided?
Answer
: The characteristic polynomial is
λ
2

4
λ
+3 = (
λ

1)(
λ

3),
thus
a
n
=
c
1
+
c
2
3
n
.
(d)
(3 points)
Find a nonhomogeneous linear recurrence whose general
solution is
a
n
=
c
1
2
n
+
c
2
(

1)
n

5.
Answer
: The characteristic polynomial should have roots 2 and

1,
hence it should be (
λ

2)(
λ
+ 1) =
λ
2

λ

2 and the homogeneous
part of the recurrence should be
a
n
+2

a
n
+1

2
a
n
= 0. Therefore, we
look for a recurrence of form
a
n
+2

a
n
+1

2
a
n
=
f
(
n
). Furthermore,
a
n
=

5 should be a solution, thus
f
(
n
) = (

5)

(

5)

2(

5) = 10.
The reccurence is
a
n
+2

a
n
+1

2
a
n
= 10.
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2
Question 2 (11 points)
Solve the following recurrences:
(a)
(3 points)
a
n
= 5
a
n
−
1
,
a
0
= 1.
Answer
:
a
n
= 5
n
.
(b)
(4 points)
a
n
+1

4

5
a
n
= 0 (find the general solution).
Answer
: We guess a solution for the nonhomogeneous recurrence as
a constant,
a
n
=
k
. We get
k

4

5
k
= 0, thus
k
=

1. The general
solution is
a
n
=
c
5
n

1.
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 Spring '09
 MarniMishna
 Characteristic polynomial, Recurrence relation, general solution

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