Math 55: Discrete Mathematics
UC Berkeley, Fall 2011
Homework # 3, due Wedneday, February 8
February 8, 2012
2.4.
26
For each of these lists of integers, provide a simple formula or rule that
generates the terms of an integer sequence that begins with the given
list. Assuming that your formula or rule is correct, determine the next
three terms of the sequence. We will refer to the terms as
a
1
, a
2
, a
3
, ...
a) 3
,
6
,
11
,
18
,
27
,
38
,
51
,
66
,
83
,
102
, ...
The pattern is
a
n
+1
=
a
n
+ 2
n
+ 1.
The next three terms are
123
,
146
,
171
.
b) 7
,
11
,
15
,
19
,
23
,
27
,
31
,
35
,
39
,
43
, ...
The pattern is
a
n
+1
=
a
n
+ 4 (which gives the formula
a
n
=
4
n
+ 3). The next three terms are 47
,
51
,
55.
c) 1
,
10
,
11
,
100
,
101
,
110
,
111
,
1000
,
1001
,
1010
,
1011
, ...
The pattern is running through all binary strings in order. (Equiv
alently: it is the sequence of natural numbers, written in binary.)
The next three terms are 1100, 1101, 1110.
d) 1
,
2
,
2
,
2
,
3
,
3
,
3
,
3
,
3
,
5
,
5
,
5
,
5
,
5
,
5
,
5
, ...
This sequence is made up of Fibonacci numbers, starting with one
1, then three 2’s then five 3’s, and so on, increasing the number
of copies by two each time. The next three terms are 8
,
8
,
8.
e) 0
,
2
,
8
,
26
,
80
,
242
,
728
,
2186
,
6560
,
19682
, ...
The pattern is
a
n
+1
= 3
a
n
+ 2. The next three terms are 59048,
177146, and 531440.
f) 1
,
3
,
15
,
105
,
945
,
10395
,
135135
,
2027025
, ...
The pattern is
a
n
+1
= (2
n
+ 1)
·
a
n
(which gives the formula
a
n
= (2
n

1)
·
(2
n

3)
·
...
·
5
·
3
·
1). The next three terms are
34459425
,
654729075
,
13749310575.
1
g) 1
,
0
,
0
,
1
,
1
,
1
,
0
,
0
,
0
,
0
,
1
,
1
,
1
,
1
,
1
, ...
The pattern is one 1, two 0’s, three 1’s, four 0’s, and so on. The
next three terms are 0
,
0
,
0.
h) 2
,
4
,
16
,
256
,
65536
,
4294967296
, ...
The pattern is
a
n
+1
=
a
2
n
. The
next three terms are
18446744073709551616
340282366920938463463374607431768211456
115792089237316195423570985008687907853269984665640564039457584007913129639936
2.4.
32
Compute the value of each of these sums.
a)
8
X
j
=0
(1 + (

1)
j
) =1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 = 5
b)
8
X
j
=0
(3
j

2
j
) =
8
X
j
=0
3
j

8
X
j
=0
2
j
=
3
9

1
3

1

2
9

1
2

1
=9330
.
c)
8
X
j
=0
(2
·
3
j
+ 3
·
2
j
) =2
·
8
X
j
=0
3
j
+ 3
·
8
X
j
=0
2
j
=2
·
3
9

1
3

1
+ 3
·
2
9

1
2

1
=21215
.
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