n j=m aj =
n i=m ai =
n k=m ak. Here, the index of summation
runs through all integers starting with
itslower limit m and ending with its upper
limit n. A large uppercase Greek letter
sigma, , is used to denote summation. The
usual laws for arithmetic app

odd integer twice d) the sequence whose
nth term is n! 2n e) the sequence that
begins with 3, where each succeeding
term is twice the preceding term f ) the
sequence whose first term is 2, second
term is 4, and each succeeding term is the
sum of the two p

5 j=1 j 2 but want the index of summation
to run between 0 and 4 rather than from 1
to 5. To do this, we let k = j 1. Then the
new summation index runs from 0
(because k = 1 0 = 0 when j = 1) to 4
(because k = 5 1 = 4 when j = 5), and the
term j 2 becomes

1, 1. Solution:(a) We recognize that the
denominators are powers of 2. The
sequence with an = 1/2n, n = 0, 1, 2, . is a
possible match. This proposed sequence
is a geometric progression with a = 1 and
r = 1/2. (b) We note that each term is
obtained by add

sequence in order of increasing
subscripts. EXAMPLE 1 Consider the
sequence cfw_an, where an = 1 n . The list of
the terms of this sequence, beginning
with a1, namely, a1, a2, a3, a4,., starts
with 1, 1 2 , 1 3 , 1 4 ,. P1: 1 CH02-7T
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2187, 6561, 19683, 59049,. n! 1, 2, 6, 24,
120, 720, 5040, 40320, 362880,
3628800,. fn 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,
89,. same recurrence relation as the
Fibonacci sequence, but with different
initial conditions). This sequence is
known as the Lucas s

k = 0 xk, |x| < 1 1 1 x
k = 1 kxk1, |x| < 1 1 (1 x)2 EXAMPLE 22 What is the value of s
cfw_0,2,4 s? Solution: Because s cfw_0,2,4 s
represents the sum of the values of s for
all the members of the set cfw_0, 2, 4, it
follows that
s cfw_0,2,4 s = 0 + 2

these sums, where S = cfw_1, 3, 5, 7? a) j S
j b) j S j 2 c) j S (1/j ) d) j S 1 31.
What is the value of each of these sums of
terms of a geometric progression? a) 8 j =
0 3 2j b) 8 j = 1 2j c) 8 j = 2 (3)j d) 8 j =
0 2 (3)j 32. Find the value of each of

famous OEIS website); and The RockClimbing Guide to New Jersey Crags with
Paul Nick. The last book demonstrates his
interest in rock climbing; it includes more
than 50 climbing sites in New Jersey. P1: 1
CH02-7T Rosen-2311T MHIA017-Rosenv5.cls May 13, 201

add 1, then multiply by 1, then add 2, then
multiply by 2, and so on h) the sequence
whose nth term is the largest integer k
such that k! n 7. Find at least three
different sequences beginning with the
terms 1, 2, 4 whose terms are generated
by a simple f

involving 3n. Comparing these terms with
the corresponding terms of the sequence
cfw_3n, we notice that the nth term is 2 less
than the corresponding power of 3. We
see that an = 3n 2 for 1 n 10 and
conjecture that this formula holds for all
n. We will se

of these series requires the use of
calculus, but sometimes they arise in
discrete mathematics, because discrete
mathematics deals with infi- nite
collections of discrete elements. In
particular, in our future studies in discrete
mathematics, we will find

Functions, Sequences, Sums, and Matrices
At each iteration of the recurrence
relation, we obtain the next term in the
sequence by adding 3 to the previous
term. We obtain the nth term after n 1
iterations of the recurrence relation.
Hence, we have added 3

n j=0 arj = arn+1 a r 1 if r =
1 (n + 1)a if r = 1. Proof: Let Sn =
n j=0 arj . P1: 1 CH02-7T Rosen-2311T
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10:24 2.4 Sequences and Summations 165
To compute S, first multiply both sides of
the equality by r and then manipu

derived in many different ways (for
example, see Exercises 37 and 38). Also
note that each of these formulae, once
known, can easily be proved using
mathematical induction, the subject of
Section 5.1. The last two formulae in the
table involve infinite se

factory. 22. An employee joined a
company in 2009 with a starting salary of
$50,000. Every year this employee
receives a raise of $1000 plus 5% of the
salary of the previous year. P1: 1 CH02-7T
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13, 2011 10:24 2.4 Seque

n j=0 arj+1 by the distributive property =
n+1 k=1 ark shifting the index of
summation, with k = j + 1 =
n k=0 ark + (arn+1 a) removing k = n +
1 term and adding k = 0 term = Sn +
(arn+1 a) substituting S for summation
formula From these equalities, we se

4 i=1 6i = 6 + 12 + 18 + 24 = 60. We can
also use summation notation to add all
values of a function, or terms of an
indexed set, where the index of
summation runs over all values in a set.
That is, we write
s S f (s) to represent the sum of the
values f

sequence (after all, there are infinitely
many different sequences that start with
any finite set of initial terms), knowing
the first few terms may help you make an
educated conjecture about the identity of
your sequence. Once you have made this
conjectu

is an equation that expresses an in terms
of one or more of the previous terms of
the sequence, namely, a0, a1,.,an1, for
all integers n with n n0, where n0 is a
nonnegative integer. A sequence is called
a solution of a recurrence relation if its
terms sa

identity here. Such a proof can be
constructed using mathematical
induction, a proof method we introduce in
Chapter 5. The proof also uses the
commutative and associative laws for
addition and the distributive law of
multiplication over addition.) We give

solution of the recurrence relation.
Many methods have been developed for
solving recurrence relations. Here, we will
introduce a straightforward method
known as iteration via several examples.
In Chapter 8 we will study recurrence
relations in depth. In

$1000 in an account that yields 9%
interest compounded annually. a) Set up a
recurrence relation for the amount in the
account at the end of n years. b) Find an
explicit formula for the amount in the
account at the end of n years. c) How
much money will t

k = 0 xk = 1 1 x , from Example 24 we
find that
k = 1 kxk1 = 1 (1 x)2 . (This
differentiation is valid for |x| < 1 by a theorem about infinite
series.) Exercises 1. Find these terms of the
sequence cfw_an, where an = 2 (3)n + 5n.
a) a0 b) a1 c) a4 d) a5

The sequences cfw_sn with sn = 1 + 4n and
cfw_tn with tn = 7 3n are both arithmetic
progressions with initial terms and
common differences equal to 1 and 4,
and 7 and 3, respectively, if we start at n
= 0. The list of terms s0, s1, s2, s3,. begins
with 1,

twentieth century. Since that time, they
have spurred a tremendous amount and
variety of research. Although many of
these problems have now been solved,
several remain open, including the
Riemann hypothesis, which is part of
Problem 8 on Hilberts list. Hi

100 k = 1 k2
49 k = 1 k2. Using the formula n k = 1k2 =
n(n + 1)(2n + 1)/6 from Table 2 (and
proved in Exercise 38), we see that
100 k = 50 k2 = 100 101 201 6 49
50 99 6 = 338,350 40,425 = 297,925.
P1: 1 CH02-7T Rosen-2311T
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cfw_an a solution of the recurrence relation
an = 8an1 16an2 if a) an = 0? b) an =
1? c) an = 2n? d) an = 4n? e) an = n4n? f )
an = 2 4n + 3n4n? g) an = (4)n? h) an =
n24n? 14. For each of these sequences
find a recurrence relation satisfied by this
seque

arrives at a hotel with a finite number of
rooms, and all rooms are occupied, this
guest cannot be accommodated without
evicting a current guest. However, we can
always accommodate a new guest at the
Grand Hotel, even when all rooms are
already occupied,

advanced mathematics, no known proof is
easy to explain. Consequently, we omit a
proof here. We refer the interested reader
to [AiZiHo09] and [Ve06] for a proof. This
result is called the Schrder-Bernstein
theorem after Ernst Schrder who
published a flawe