11.1 sequences sequence terms,
TRANSCRIPT
11.1 Sequences
Informally speaking a sequence is a succession of numbers, called terms, in a definite order
Example: ,4
1,
3
1,
2
1,1
1 is the first term of the sequence; 2
1 is the second term;
3
1 is the third terms and so on
Here is the formal definition of a sequence
Definition
A sequence is a function whose domain is the set of natural numbers {1,2,3,…}
A sequence of real number has as the range the subset of real numbers. A sequence function assigns to each
positive integer n a real number, that we denote an (instead of f(n)). Sequences are traditionally denoted by
{an}.
A sequence can be defined by
- Listing first few terms. Enough terms must be listed so it is clear how the terms of the sequence are
formed
Example: ,4
1,
3
1,
2
1,1
It should be clear here that the next terms are ,7
1,
6
1,
5
1
- Giving a formula for the n-th term
Example: 1,2
1 na
nn , or 12
1
n
n
Formula allows us to compute any term of the sequence. If we want 10-th term, or a10, we replace n by
10 in the formulanna
2
1 :
1024
1
2
11010 a
- A recursive formula where we define explicitly a first (or the first few) term(s) of a sequence and define
the n-th term by an equation that involves the preceding terms
Example: a1 = 3, an = 2an-1+ 5 for n >1.
We can compute successive terms by using the formula. So a2 = 2a2-1+5 =2a1+5 =23+5 = 11 ; a3 = 2a3-1+5
=2a2+5 =211+5=27, and so on.
Example:
Write down the first five terms of the sequence {bn} = {n2-1}
b1=12-1=0; b2 = 22 -1=3 ; b3=32-1= 8 ; b4= 42 -1=15; b5 = 52-1 = 24
Therefore the sequence is : 0, 3, 8, 15, 24, ….
Example:
Write the first five terms of the sequence defined recursively as a1 = 1, a2 = 2, an = 2an-1 + an-2
We already have a1 and a2. We’ll use formula an = 2an-1 + an-2 for n = 3, 4, 5
a3 = 2a3-1 + a3-2= 2a2+a1=22+1=5
a4 = 2a4-1 + a4-2=2a3+a2= 25+2= 12
a5 = 2a5-1 + a5-2= 2a4+a3 = 212+5=29
Therefore the sequence is: 1, 2, 5, 12, 29, ….
Having a formula for the n-th term of a sequence is the most efficient way to work with a sequence, since we
can quickly compute any term of that sequence. For example, if we need 100-th term of a sequence defined
recursively we would most likely have to compute all 99 preceding terms. However, not always such a formula
is easy to establish.
Example:
Find the formula for the n-th term of a sequence an for which the first five terms are
16
1,
8
1,
4
1,
2
1,1
Note that each term is a fraction with the numerator 1 and the denominators are consecutive powers of 2:
,2
1
8
1,
2
1
4
1,
2
1
2
1,
2
11
3210
Therefore ,2
1
8
1,
2
1
4
1,
2
1
2
1,
2
11
34231201 aaaa . Notice that in the n-th term
(first, second, third…) the powers of 2 are n-1. So, we conclude that ...3,2,1,2
11
nann . We could simplify
the formula, if we started counting from zero: ,...3,2,1,0,2
1 na
nn
Example:
Find the formula for the n-th term of a sequence an for which the first five terms are: 1, -1, 1, -1, 1 …
In this sequence, the terms alternate the sign. Note that (-1)even power = 1 and (-1)odd power= - 1. Since for positive
integers, the numbers alternate between odd and even, we see that {(-1)n-1}n >1 ={1, -1, 1, -1, …} and
{(-1)n}n >1 ={-1, 1, -1, 1, …}. Therefore, for the given sequence an = (-1)n-1 , n = 1,2,3,…
Geometrically a sequence can be illustrated in two ways
- The terms of a sequence can be plotted on the number line
- Points (n, an) can be plotted in the coordinate plane and represent the graph of the function (a(n))
Example:
Illustrate the sequence ,4
1,
3
1,
2
1,1 on the number line.
The sequence ,4
1,
3
1,
2
1,1 can be represented by a formula an =
n
1, n > 1. Numbers an are plotted on the
number line:
Example:
Illustrate the sequence {(-1)nn} in the coordinate plane.
Points (n, (-1)nn) are plotted in the coordinate plane. The first few points are (1,-1), (2,2), (3,-3), (4,4)…
Sigma notation
If a sequence {an} is given then the sum of the first n consecutive terms of such sequence is naaaa 321
are shortly written using sigma notation. Sigma is a Greek letter which corresponds to S in English alphabet. It is
written as . So the sum above is written as
n
k
kn aaaaa1
321
In the notation
n
k
ka1
, ka indicates that we must add the consecutive terms of a sequence {ak}; the “k=1”on
the bottom of tells us to start with the term ak, where k = 1 (that is with the term a1); the “n” on the top of
, tells us to stop adding when we reach term an. For example, 54321
5
1
aaaaaak
k
. We can use
sigma notation to indicate any sum of consecutive terms of a sequence . For example,
1098765
10
5
aaaaaaak
k
Example:
Write out (expand ) the sum
5
1
)32(k
k and find its value.
We must add consecutive terms of the sequence ak = 2k+3, starting with k=1 and ending with k = 5
451311975)352()342()332()322()312()32(5
1
k
k
Example:
Express the sum using sigma notation 729
64
243
32
81
16
27
8
9
4
3
2
First note that
729
64
243
32
81
16
27
8
9
4
3
2
729
64
243
32
81
16
27
8
9
4
3
2
Therefore we must add 6 terms of the sequence:
,729
64,
243
32,
81
16,
27
8,
9
4,
3
2
First we find the formula for n-th term of this sequence. Note that the numerators are powers of 2 and the
denominators are powers of 3. The sign alternates, so there must be (-1)n or (-1)n+1 present. Therefore,
1,3
2)1(
3
2)1( 11
na
n
n
n
nn
n. We use this term within sigma:
k
k
k
6
1
1
3
2)1(
729
64
243
32
81
16
27
8
9
4
3
2
729
64
243
32
81
16
27
8
9
4
3
2
Properties of sigma notation
If {ak} and {bk} are two sequences of real numbers then
(I)
m
nk
m
nk
kk acca
(II)
m
nk
m
nk
m
nk
kkkk baba )(
Property (I) is simply generalized distribution property, and property (II) follows from the fact that
when adding numbers we can group them in any way we like (commutative and associative properties
of addition)
Example:
Suppose that {ak} and {bk} are two sequences such that
15
1
15
1
6,9k
k
k
k ba
Then
15
1
15
1
15
1
15
1
15
1
39)6(2932323)23(k k
k
k
kk
k k
kkk bababa
Special Sums
(A)
n
k termsn
cnccccc1
... , where c is any real number
(B)
n
k
nnnk
1 2
)1(...321
(C)
n
k
nnnnk
1
22222
6
)12)(1(...321
These formulas can be proven in an elementary way. We omit the proof here.
Properties of sigma and special sum allow us to compute large sums with a relative ease.
Example:
Find the value of the sum
184
1
)53(k
k
184
1
184
1
184
1
184
1
980,519201859239202
)1184(18431845353)53(
k kk k
kkk
11.2 Arithmetic Sequences
A sequence }{ na is arithmetic, if the difference of two consecutive terms is always the same, that is for any
n > 1, daa nn 1 =constant (independent of n). The difference, d, is called the common difference.
Note that the terms of an arithmetic sequence are formed by adding the common difference, d, to the previous
term: daa nn 1 . Therefore, if a1 is the first term of the arithmetic sequence, then,
...
3)2(
2)(
1134
1123
12
daddadaa
daddadaa
daa
Example:
Determine whether given sequence is arithmetic.
a)
2
1}{
n
nan , n > 1
We must show that 1 nn aa is constant, for n > 1. Since 2
1
n
nan , then
12)1(
1)1(1
n
n
n
nan
Therefore )1)(2(
1
)1)(2(
212
)1)(2(
)2()1)(1(
12
1 22
1
nnnn
nnnn
nn
nnnn
n
n
n
naa nn .
The difference 1 nn aa depends on n, so it is not a constant, and therefore the sequence is not arithmetic.
b) }13{}{ nbn , n >1
We must show that 1 nn bb
is a constant, for n > 1. Since 13 nbn , then 231)1(31 nnbn
Therefore 32313)23()13(1 nnnnbb nn . Since 1 nn bb is a constant, sequence is
arithmetic.
Theorem
If }{ na is an arithmetic sequence with the first term a1 and common difference d, then
dnaan )1(1 , n > 1
and
naa
aaaaan
k
n
kn
1
1
3212
Example:
Find the 26-th term of an arithmetic sequence with the first term ½ and the common difference -1.
Denote the sequence }{ na . Then a1= ½ and d = -1. Since the sequence is arithmetic,
2
3)1)(1(
2
1)1(1 nndnaan for n > 1. Therefore , when n = 26, we get
2
49
2
32626 a
Example:
Find the 200-th term of the arithmetic sequence ,37,35,33,3
Denote the sequence }{ na . Since the sequence is arithmetic, 32333,3 121 aada .
Therefore, the n-th term of the sequence is
3)12(332)32)(1(3)1(1 nnndnaan
And consequently, 33993)12002(200 a
Example:
Find the first term and the common difference, d, for an arithmetic sequence whose 4th term is 3 and 20th term
is 35. Give the recursive formula for the sequence and write the formula for the n-th term
If }{ na is the sequence, then 353 204 aanda . Since the sequence is arithmetic, then dnaan )1(1
for n > 1.
Therefore,
daa
daa
1935
33
120
14
Solving the system
3519
33
1
1
da
da
for a1 and d, we get d= 2 and a1 = -3.
Therefore the n-th term of this sequence is 522)1(3)1(1 nndnaan and the recursive
formula is
2
3
11
1
nnn adaa
a
Example:
Find the following sum of the arithmetic sequence
31+ 34+ 37 + … + 88
Denote the sequence }{ na . Since the sequence is arithmetic, 33134,31 121 aada .
Then 2833)1(31)1(1 nndnaan . To apply the formula for the sum of the first n terms of an
arithmetic sequence ( naa
aaaaan
k
n
kn
1
1
3212
), we must find out how many terms of given
sequence }{ na are added or the value of n. Since the last term in the given sum is 88, then an = 88 and
therefore, 3n+28 = 88. Solving this equation gives n= 20.
Thus we get
31+ 34+ 37 + … + 88 1190202
8831
2
1
naa n
Example :
Find the sum
80
1 3
1
2
1
n
n
We can use properties and formulas discussed in the previous section to evaluate this sum or we can notice
that the sequence 3
1
2
1 nan is arithmetic and use formula for the first n terms of an arithmetic sequence.
Since the first method is easier, we’ll use properties of sigma to evaluate this sum
3
4940
3
80812080
3
1
2
)180(80
2
1
3
1
2
1
3
1
2
1 80
1
80
1
80
1
nnn
nn
11.3 Geometric sequence
A sequence }{ na is geometric , if the ratio of two consecutive terms is always the same nonzero number, that is
for any n > 1, ra
a
n
n 1
=constant (independent of n), r≠0, a1≠0 . The ratio r , is called the common ratio.
Note that the terms of a geometric sequence are formed by multiplying the previous term by the common
ration r, 1 nn raa . Therefore, if a1 ≠0, is the first term of the arithmetic sequence, then,
...
)(
3
1
2
134
2
1123
12
rararraa
rararraa
raa
Example:
Determine whether given sequence is geometric.
a)
nn
na
2
1}{ , n > 1
We must show that1n
n
a
a is a constant, for n > 1. Since
nn
na
2
1 , then
11122
1)1(
nnn
nna . Therefore
n
n
n
n
n
n
a
an
n
n
n
n
n
2
1
2
2)1(
2
2
11
11
. The quotient 1n
n
a
a depends on n, so it is not a constant, and therefore the
sequence is not geometric.
b)
1
5
32}{
n
nb , n >1
We must show that 1n
n
b
b is a constant, for n > 1. Since
1
5
32
n
nb , then
21)1(
15
32
5
32
n
nb
Therefore 5
3
5
32
5
32
1
1
n
n
n
n
b
b. Since
1n
n
b
b is a constant, sequence is geometric.
Theorem
If }{ na is a geometric sequence with the first term a1 ≠ 0 and common ratio r≠ 0, then
1
1
n
n raa , n > 1
and
n
k
n
kn
rifna
rifr
ra
aaaaa1
1
1321
1
11
1
Example:
Find the 8-th term of a geometric sequence with the first term 4 and the common ratio -1/2 .
Denote the sequence }{ na . Then a1= 4 and d = -1/2 . Since the sequence is geometric,
1
1
12
14
n
n
n raa for n > 1. Therefore , when n = 8, we get32
1
2
)1(4
2
14
7
77
8
a
Example:
Find the 50-th term of the geometric sequence ,39,9,33,3,3
Denote the sequence }{ na . Since the sequence is geometric, 33
3,3
1
21
a
ara .
Therefore, the n-th term of the sequence is 11
1 33
n
n
n raa
And consequently, 252/5050150
50 33333
a
Example:
Find the first term and the common ratio, r, for a geometric sequence whose 4th term is 15 and 20th term is 45.
Give the recursive formula for the sequence and write the formula for the n-th term
If }{ na is the sequence, then 353 204 aanda . Since the sequence is geometric , then 1
1
n
n raa
for n > 1.
Therefore,
19
120
3
14
45
15
raa
raa
To find a1 and r, we must solve the system
45
15
19
1
3
1
ra
ra
Note that 16
3
1
19
1
15
45r
ra
ra , hence, 316 r and 16
1
3316 r . Then , using the first equation, we get
163319
15
3
15
3
1515
16
3
16
1
ra
Therefore the n-th term of this sequence is 16
416
3
16
1
16
3
16
1
11
1 315)3(15
3
)3(153
3
1516
1
16
3
nn
n
nn
n raa
and the recursive formula is
116
1
1
16
31
3
3
15
nnn araa
a
Example:
Find the following sum of the geometric sequence 21
5
32
25
18
5
62
Denote the sequence }{ na . Since the sequence is geometric, 5
3
2
5
6
,21
21
a
ara .
Then
1
1
15
32
n
n
n raa . To apply the formula for the sum of the first n terms of an arithmetic sequence (
n
k
n
kn
rifna
rifr
ra
aaaaa1
1
1321
1
11
1
), we must find out how many terms of given sequence
}{ na are added that is we need the value of n. Since the last term in the given sum is
21
5
32
, then
21
5
32
na and therefore,
211
5
32
5
32
n
na . Clearly, n-1= 21, that is n = 22
Thus we get
22
22
22
1
21
5
315
5
31
5
31
21
1
5
32
25
18
5
62
r
ra
Example :
Find the sum
15
1
134
k
k
Sequence an = 4(3)n-1
is geometric since 3334
34 )2(1
1)1(
1
1
nn
n
n
n
n
a
a. Therefore, r = 3 and a1 = 4(3)1-1 =4.
Hence
)31(231
314
1
134 15
151515
1
1
15
1
1
r
raa
k
k
k
k
11.5 Binomial Theorem
The Binomial Theorem give the general formula for (x+a)n. Before stating it however, we need to introduce
notation.
Definition:
If n is a positive integer, we define n factorial, denoted as n!, as
0! = 1
1! = 1
nn 321!
Binomial coefficient
k
n is defined as
)!(!
!
knk
n
k
n
, 0 < k < n. The symbol
k
n is read “ n choose k”.
Example:
Compute 6!
720654321!6
Note that 65!4654321!6!4
In general, for any k < n, we have nkkkn )2)(1(!!
Example:
Since 5 < 9, we can rewrite 9! as 9876!5!9 .
Similarly 11! Can be written as 11109!8!11 or 11109876!5!11
Example:
Compute the following binomial coefficients :
3
7and
0
5
a) 35321
765
!4!3
765!4
!4!3
!7
)!37(!3
!7
3
7
b) 1!51
!5
!5!0
!5
)!05(!0
!5
0
5
Here are some properties of binomial coefficients
Theorem
If n, k are positive integers such that k < n, then
(i) 10
n; 1
n
n
(ii) nn
1; n
n
n
1
(iii)
kn
n
k
n
(iv)
k
n
k
n
k
n 1
1
Properties (i) –(iii) follow directly from the definition of the binomial coefficient. To show that property (iv) holds
we use the definition and algebra to transform the left hand side into the right hand side:
k
n
knk
n
kknknk
nn
kkn
knk
knk
n
kknknk
n
knkk
n
knknk
n
knk
n
knk
n
knk
n
knk
n
k
n
k
n
1
)!1(!
)!1(
)1()!()!1(
)1(!
)1(
1
)!()!1(
!
1
1
1
)!()!1(
!
)!()!1(
!
)!1()!()!1(
!
)!(!
!
)!1()!1(
!
)!(!
!
))!1(()!1(
!
1
Example
a)
4
6
4
5
3
5 (here n = 5 and k = 4)
b)
3
8
3
7
2
7 (here n = 7 and k = 3)
Binomial Theorem
nnnnnkknn
k
n aaxn
nax
nax
nxax
k
nax
11221
0 121)(
Remarks:
Note that
nnnnnnnkkn
n
k
axn
nax
n
nax
nax
nax
nax
k
n112210
0 1210
nnnnn aaxn
nax
nax
nx
11221
121
Note that in the expansion of (x+a) n, the exponents of x decrease from n to 0, while the exponents of a increase
from 0 to n. The sum of the exponents of x and a in each term is always n
Example
Use the Binomial Theorem to expand (x+2)5 . Here n = 5 and a = 2.
32808040103216581041025
25
52
4
52
3
52
2
52
1
52
0
52
5)2(
23452345
55544533522511500555
0
5
xxxxxxxxxx
xxxxxxxk
x kk
k
Example
Use the Binomial Theorem to expand (x2 - y2)6.
First we must re-write the binomial so it resembles the binomial in the Theorem: (x2 - y2)6 =(x2 +(- y2))6 . So, here
n = 6, x is replaced by x2 and a is replaced by (-y2)
1210284664821012
12102846648210126266252562
4246232362222621216202062
2626
0
622622
61520156
)(615)(2015)(6)(6
6
5
6
4
6
3
6
2
6
1
6
0
6
6))(()(
yyxyxyxyxyxx
yyxyxyxyxyxxyxyx
yxyxyxyxyx
yxk
yxyxkk
k
Example
Find the coefficient of x6 in the expansion of (x+3)10
By the Binomial Theorem
kkk
k
kkk
k
kk
k
xk
xk
xk
x
1010
10
0
101010
0
1010
0
10 3210
3210
3)2(10
)32(
The terms in the expansions are of the form kkk xk
1010 3210
. Since we are looking for the term that contains
x6, we want 10-k to be 6: 10-k = 6. Therefore k = 4. Then the coefficient is kk
k32
1010
with k = 4. Therefore,
the coefficient of x6 is 640,088,1816421032210324
10464410
Coefficients of (x+a)N can be arranged in a triangular table called Pascal’s triangle
N = 0 1
N= 1 1 1
N=2 1 2 1
N= 3 1 3 3 1
N= 4 1 4 6 4 1
N= 5 1 5 10 10 5 1
To find the coefficients of (x +a)5 start with a 1, the next coefficient will be the sum of the two coefficients in the
row above ( shown in red: 5 = 1+4) , next coefficient will be the sum of 4 and 6 (thus 10) , the next (shown in
blue) is the sum of 6 and 4, and so on. End the row with a 1 to form a triangular array of numbers.
Using Pascal’s triangle and remark about the exponents of x and a, we can quickly write the expansion of
binomial (x+a)n for small values of n. For example,
(x+a)5 = 1x5a0+ 5x4a1+10x3a2+10x2a3+5x1a4+1x0a5= x5+5x4a +10x3a2+10x2a3+5xa4+ a5