Quiz 4 Solutions
Dilip Raghavan
November 11, 2008
Problem 1.
(a) If the first slot is not a 0, then we have 3 possibilities for the first slot and
a
n

1
. If the first
slot is 0, then we nust have an odd number of 0 in what follows. The total number of sequences of
length
n

1 is 4
n

1
, and of these
a
n

1
have an even number of 0s. So there are 4
n

1

a
n

1
possibilities
for this case. So
a
n
=
3
a
n

1
+
4
n

1

a
n

1
=
2
a
n

1
+
4
n

1
.
(b) The homogeneous solution is
A
2
n
. Look for a particular solution of the form
p
(
n
)
=
B
4
n
. Substituting,
we get the equation
B
4
n
=
2
B
4
n

1
+
4
n

1
, which simplifies to 4
B
=
2
B
+
1 Hence,
B
=
1
2
.
So
a
n
=
A
2
n
+
1
2
4
n
. Using the initial condition
a
1
=
3, we get
A
=
1
2
.
Problem 2.
Let
a
n
be the number of such subsets. We find a recurrence relation for
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 Spring '11
 Raghaven
 Natural number, consecutive integers, Dilip Raghavan, an2 such subsets, an1 such subsets

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