11.1 Sequences sequence terms,

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