Solutions to the Practice Final Exam
December 2, 2009
1. Solution:
(a)
a
(0) = 1
, a
(1) = 2
, a
(2) = 7
, a
(3) = 26
, a
(4) = 97, and
a
(5) = 362.
(b) Notice that the characteristic polynomial of the recurrence relation is
x
2

4
x
+1, which has roots 2
±
√
3,
yielding template solution
a
(
n
) =
α
(2 +
√
3)
n
+
β
(2

√
3)
n
. From the initial values, we have
α
+
β
=
1
α
(2 +
√
3) +
β
(2

√
3)
=
2
which has solution
α
= 1
/
2 and
β
= 1
/
2. Thus, we have
a
(
n
) =
(2 +
√
3)
n
2
+
(2

√
3)
n
2
.
(c)
Proof by induction.
Step 1. (Base case)
It is easy to verify that
a
(0) =
(2+
√
3)
0
2
+
(2

√
3)
0
2
= 1 and
(2+
√
3)
1
2
+
(2

√
3)
1
2
= 2.
Step 2. (Induction)
Suppose
a
(
n
) =
(2+
√
3)
n
2
+
(2

√
3)
n
2
for some
n
≥
1. We then compute
a
(
n
+ 1)
=
4
a
(
n
)

a
(
n

1)
=
4(
(2 +
√
3)
n
2
+
(2

√
3)
n
2
)

(
(2 +
√
3)
n

1
2
+
(2

√
3)
n

1
2
)
=
(2 +
√
3)
n

1
2
(4(2 +
√
3)

1) +
(2

√
3)
n

1
2
(4(2

√
3)

1)
=
(2 +
√
3)
n

1
2
(2 +
√
3)
2
+
(2

√
3)
n

1
2
(2

√
3)
2
=
(2 +
√
3)
n
+1
2
+
(2

√
3)
n
+1
2
and the induction step is complete.
2. Solution:
Notice that
x
(
n
+ 1) +
n
+ 1 = 2(
x
(
n
) +
n
) + 1. Letting
y
(
n
) =
x
(
n
) +
n
, we have
y
(
n
+ 1) = 2
y
(
n
) + 1
, n
≥
1
.
Since
x
(1) = 1, then
y
(1) = 2.
Guessing now that the general solution has the form
y
(
n
) =
α
2
n
+
β
, if we plug into the recurrence we get
α
2
n
+1
+
β
= 2
·
α
2
n
+ 2
β
+ 1
which implies
β
=

1. Now substituting into the recurrence for
n
= 1, we get
α
=
3
2
. Thus,
y
(
n
) = 3
·
2
n

1

1
which implies that
x
(
n
) = 3
·
2
n

1

n

1
.
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 Fall '07
 Graham
 1 digit, 1 k, 2k, 2 K, 2 digits

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