Sequences and SeriesCHAPTER 1

Chapter 1: Sequences and Series

1.1 – ARITHMETIC SEQUENCES

CREATE YOUR OWN

Using the graph paper provided, create your own Fibonacci Spiral, like Vi Hart.

Instructions (if you need them):• Begin in the right bottom third of the page, and draw a 1 cm

x 1 cm square.• Below the first square, draw another 1 cm by 1 cm square.

To the left of these squares, draw a 2 cm by 2 cm square.• Below these, draw a 3 cm by 3 cm square.• Carry on.• Use the diagonals between the corners to guide your curve.

NOTATION

Say we wanted to represent the Fibonacci Sequence as follows:• 1 = t1

• 1 = t2

• 2 = t3

• Etc…

• What would notation would we use for the 4th term? For the 7th?

• What notation would we use for the nth term? • tn

• How about the term before the nth term? • tn-1

• How about the term two before the nth term? • tn-2

• Can you make a general formula for tn in the Fibonacci Sequence?• tn = tn-1 + tn-2

SOME DEFINITIONS

A sequence is an ordered list of objects. It contains elements or terms that follow a pattern or rule to determine the next term in the in the sequence.

The first term of the sequence is t1.The number of terms in the sequence is n.The general term of the sequence is tn. A finite sequence always has a finite number

of terms. ex: 1, 2, 3, 4, 5

2, 4, 6, 8, …, 42An infinite sequence never ends.

ex: 2, 4, 6, 8, …An arithmetic sequence is a sequence in which there is a common difference between consecutive terms (i.e. the difference between terms is constant).

CONSIDER THE SEQUENCE:10, 16, 22, 28, …

Terms t1 t2 t3 t4 t6

Sequence

10 16 22 28 34

• What’s the common difference?• What’s the next term?

t1 =t2 =t3 =…tn =

t1

t1 + dt1 + 2d

t1 + (n-1)d

EXAMPLE

A visual and performing arts group wants to hire a community events leader. The person will be paid $12 for the first hour of work, $19 for two hours of work, $26 for the three hours of work, and so on.

a) Write the general term that you could use to determine the pay for any number of hours worked.

b) What will the person get paid for 6h of work?

t1 = 12t2 = 19t3 = 26

What’s the common difference?

19 – 12 = 726 – 19 = 7d = 7

a) General term is always in the form:

tn = t1 + (n – 1)d

tn = 12 + (n – 1)7 tn = 12 + 7n – 7 tn = 7n + 5

b) We are looking for the sixth term, t6.

n = 6 t6 = 7(6) + 5 t6 = 42 + 5 t6 = 47

The general term is:

tn = 7n +5

They will get paid $47 for 6 hours of work.

ON YOUR CALCULATOR

Let’s look at the same series:

t1 = 12t2 = 19t3 = 26tn = 7n + 5

The formula for the general term can be treated just like a function. So, we can graph the expression 7n+5 in our Y= on our graphing calculator, and use our graph to solve for n=6.

Instructions:

Y= 7 x + 5 GRAPH TRACE 6

Now, if you wanted to find any term, you could just hit TRACE and the NUMBER, and it will give you your answer in the “y=“

You can see, to the right of your screen, it says y = 47, which was our answer.

EXAMPLE

The musk-ox of northern Canada are hoofed mammals that survived the Pleistocene Era, which ended 10 000 years ago. In 1955, the Banks Island musk-ox population was approximately 9250 animals. Suppose that in subsequent years, the growth of the musk-ox population followed an arithmetic sequence, in which the number of musk-ox increased by approximately 1650 each year. How many years would it take for the musk-ox population to reach 100 000?What variable will 100 000 represent?What are we looking for?

t1 = 9250d = 1650tn = 100 000n = ?

9250, 10 900, 12 550, …, 100 000

tn = t1 + (n – 1)d

100 000 = 9250 + (n – 1)1650 100 000 = 9250 + 1650n – 1650 100 000 – 9250 + 1650 = 1650n 92 400 = 1650n 56 = n

It would take 56 years for the musk-ox population to reach 100 000.

TRY IT!

t3 = 16t8 = 6 t9 = 4

a)d = 4 – 6 = -2t1 = 16 + 2 + 2t1 = 20

b) tn = t1 + (n – 1)(d)

tn = 20 + (n – 1)(-2) tn = 20 – 2n + 2 tn = —2n + 22

c) We’re told that the 8th row has six boxes, so we can see that’s third from the top. So, there must be 10 rows.

Algebraically:2 = —2n + 22 —20 = —2n n = 10

Independent practice

PG. 16–20 #4, 5, 8, 12, 14, 17, 20, 21,

23, 24

Chapter 1: Sequences and Series

1.2 – ARITHMETIC SERIES

HANDOUT

The handout follows the activity outlined on page 22 of your textbook. Answer all the questions to your fullest ability, as this is a summative assessment.

ARITHMETIC SERIES

An arithmetic series is the sum of terms that form an arithmetic sequence. For the arithmetic sequence 2, 4, 6, 8, the arithmetic series is represented by 2 + 4 + 6 + 8.

We can express an arithmetic series as follows:Sn = t1 + (t1 + d) + (t1 + 2d) + … + (t1 + (n – 3)d) + (t1 + (n – 2)d) + (t1

+ (n-1)d)

Recall: t1 =t2 =t3 =…tn =

t1

t1 + dt1 + 2d

t1 + (n-1)d

How can we use Gauss’ method to create a formula for the sum of an arithmetic series?

FORMULA

We can express an arithmetic series as follows:Sn = t1 + (t1 + d) + (t1 + 2d) + … + (t1 + (n – 3)d) + (t1 + (n – 2)d) + (t1

+ (n-1)d)Sn = t1 + (t1 + d) + … + (t1 + (n – 2)d) + (t1 + (n-1)d)Sn = [t1 + (n-1)d] + [t1 + (n – 2)d] + … + (t1 + d) + t1 2Sn = [ 2t1 + (n – 1)d] + [2t1 + (n – 1)d] + … + [2t1 + (n – 1)d] + [2t1

+ (n – 1)d]

2Sn = n[2t1 + (n – 1)d]

Sn = (n/2)[2t1 + (n – 1)d]

FORMULA CONTINUED

Alternatively, we can substitute tn into the formula to find:

So, the two formulas for the sum of arithmetic series are:

EXAMPLE

The sum of the first two terms of an arithmetic series is 13 and the sum of the first four terms is 46. Determine the first six terms of the series and the sum to six terms.

Given:S2 = 13S4 = 46

Solve the system of equations:1

2

1 – 2

Substitute d = 5 into one of the equations:

So, the first six terms are 4, 9, 14, 19, 24, 29.

andS6 = (6/2)(4+29) = 3(33) = 99

TRY IT!

The sum of the first two terms of an arithmetic series is 19 and the sum of the first four terms is 50. What are the first six terms of the series and the sum to 20 terms?Given:S2 = 19S4 = 50

System of Equations:

The first six terms are 8, 11, 14, 17, 20, 23.

S20 = (20/2)(8 + 23) = 10(31) = 310

Independent practice

PG. 27 – 31, # 1, 6, 9, 12, 14, 16-20, 23

Chapter 1: Sequences and Series

1.3 – GEOMETRIC SEQUENCES

GEOMETRIC SEQUENCES

A geometric sequences is a sequence in which the ratio of consecutive terms is constant. The Fibonacci Sequence is an example of a geometric sequence.

Do the activity described on page 33. Answer questions # 1 – 4, and create a table like that seen in question #2.

The common ratio of a geometric sequence is the ratio of successive terms in that sequence.

TABLE FROM COIN TOSS ACTIVITY

Number of Coins, n

Number of Outcomes, tn

Expanded Form

Using Exponents

1 2 (2) 21

2 4 (2)(2) 22

3 8 (2)(2)(2) 23

4 16 (2)(2)(2)(2) 24

…

n 2n (2)(2)…(2) 2n

FORMULA

We can always find r, the common ratio, by dividing one term by the one preceding it, or by using this formula:

Terms of a geometric sequence:

t1 = t1

t2 = t1rt3 = t1r2

t4 = t1r3

…tn = t1rn-1

The general term for a geometric sequence where n is a positive integer is:

tn = t1rn-1

EXAMPLE

Sometimes you use a photocopier to create enlargements or reductions. Suppose the actual length of a photograph is 25 cm and the smallest size that copier can make is 67% of the original. What is the shortest possible length of the photograph after 5 reductions? Express your answer to the nearest tenth of a centimetre.

t1 = r = n =

tn = t1rn-1

t6 = (25)

(0.67)5

t6 = 3.375…

After five reductions, the shortest possible length of the photograph is approximately 3.4 cm.

25 cm0.676

EXAMPLE

In a geometric sequence, the third term is 54 and the sixth term is -1458. Determine the values of t1 and r, and list the first three terms of the sequence.

t3 = 54t6 = -1458

t3 = 54t4 = 54rt5 = 54r2

t6 = 54r3 = -1458

t6 = 54r3 = -1458

r3 = -1458/54 = -27r = -3

t2 = 54/r = 54/-3 = -

18

t1 = -18/-3 = 6

t1 = 6The first three terms are 6, -18, 54.

Method 1: Method 2:

tn = t1rn-1

54 = t1r2

t1 = 54/r2

Substitute: -1458 = t1r5

r5 = -1458/(54/r2) r5 = -1458(r2/54) r5/r2 = -27 r3 = -27 r = -3

From there it’s the same!

EXAMPLE

The modern piano has 88 keys. The frequency of the notes ranges from A0, the lowest note, at 27.5 Hz, to C8, the highest note on the piano, at 4186.009 Hz. The frequencies of these notes approximate a geometric sequence as you move up the keyboard.a) Determine the common ratio of the geometric sequence produced from

the lowest key, A0, to the fourth key, C1, at 32.7 Hz.b) Use the lowest and highest frequencies to verify the common ratio

found in part a).a) t1 = 27.5 Hz n = 4 tn = 32.7 Hz

tn = t1rn-1

32.7 = (27.5)r3

r3 = 1.189… r = 1.0594…

The common ratio is approximately 1.06.

b) t88 = t1r88-1

4186.009 = (27.5)r87

r87 = 152.218… r = 1.0594…

It’s the same!

Do you know how to find the 87th root of a number on your calculator?

Independent practice

PG. 39–45, # 1, 5, 9, 10, 13, 15, 16, 19, 21, 23, 24, 27

Chapter 1: Sequences and Series

1.4 – GEOMETRIC SERIES

GEOMETRIC SERIES

Considering what you know about arithmetic series, do you think a geometric series might be?

A geometric series is the terms of a geometric sequence expressed as a sum.

The formula for the sum of a geometric series is:

EXAMPLE

Determine the sum of the first 10 terms of each geometric series.a) 4 + 12 + 36 + …b) t1 = 5, r = 1/2

a) In the series, t1 = 4, r = 3, and n = 10.

b) t1 = 5, r = ½, n = 10

EXAMPLE

Determine the sum of each geometric series.a) 1/27 + 1/9 + 1/3 + … + 729

a) tn = t1rn-1

729 = (1/27)(3)n-1

(27)(729)=3n-1

(33)(36) = 3n-1

39 = 3n-1

9 = n – 1n = 10

There are ten terms in this series.

The sum of the series is 29524/27.

ALTERNATE FORMULA METHOD

Determine the sum of each geometric series.a) 1/27 + 1/9 + 1/3 + … + 729

We can make an alternate formula that can make these kind of problems one step!

The general formula can also be written like this.

Using the general formula for a geometric sequence, we can multiply each side by r to find that t1rn can be equal to rtn

Substitute:

Our series:

REVISED FORMULA

EXAMPLE

The Western Scrabble Network is an organization whose goal is to promote the game of Scrabble. It offers Internet tournaments throughout the year that WSN members participate in. The format of these tournaments is such that the losers of each round are eliminated from the next round. The winners continue to play until a final match determines the champion. If there are 256 entries in an Internet Scrabble tournament, what is the total number of matches that will be played in the tournament?The players match up and play against each other, so for the first round there are 256/2 = 128 matches = t1. Each round, half the players are eliminated, so r = ½. At the end of the tournament there will be only one final match, so tn = 1.

There will be 255 matches played at the tournament.

Independent practice

PG. 53–57, # 4, 6, 8, 10, 11, 13, 15, 16-

19, 22

Chapter 1: Sequences and Series

1.5 – INFINITE GEOMETRIC SERIES

ZENO’S PARADOX

PULL OUT A SHEET OF PAPER

1. Draw a line dividing your page in two halves (fold it if necessary). Shade in one half.

2. Divide the non-shaded half into two, and shade one of those halves.3. Divide the new non-shaded half into two, and shade one of those

halves. Continue on at least six more times.

What’s the fraction of each part?

1/2

1/4

1/8

1/16

1/32 So, the first five terms are:1/2, 1/4, 1/8, 1/16, 1/32 …

What will the next two terms be?What’s the general term?

tn = (1/2)(1/2)n-1 = (1/2)n

GRAPH IT

tn = (1/2)n• Enter (1/2)x into your Y= on your

calculator.• What does the graph look like?• What happens to the y-value as the x-

value gets larger?What will our the sum of this geometric series look like?

• Enter 1 – (1/2)x into your Y=• What does the graph look like? Look at the

table of values.• What happens as the x-value gets larger?

What’s it getting closer to?

Vi Hart on Infinite Series

More?

CONVERGENT SERIES

Consider the series 4 + 2 + 1 + ½ + …

Graph it! Put —8[(1/2)x – 1] into your Y= and check the graph.

A series is a convergent series when it has an infinite number of terms, in which the sequence of partial sums approaches a fixed value.

We can tell this series is convergent by looking at its graph and table of values, and seeing that it is approaching the value of 8.

DIVERGENT SERIES

Consider the series 4 + 8 + 16 + 32 + …

Graph it! Put 4(2x – 1) into your Y= and check the graph.

A series is a divergent series when it has an infinite number of terms, in which the sequence of partial sums does not approach a fixed value.

We can see from looking at this graph and the table of values that the y-value is not approaching any number—it’s just getting bigger.

FORMULA

We can re-arrange for the formula for a finite geometric series to find:

Why might this be helpful when r < 1?

So, if r < 1, then as n gets very large, what will happen to rn?rn gets closer to zero as n gets larger for r < 1.

So, for an infinite series, where n is getting larger and larger, approaching infinity, what can we say about (1 – rn)? What can we say about Sn?

For an infinite series: where -1 < r < 1

EXAMPLE

Assume that each shaded square represents ¼ of the area of the larger square bordering two of its adjacent sides and that the shading continues indefinitely in the indicated manner.

a) Write the series of terms that would represent this situation.b) How much of the total area of the largest square is shaded?

a) What portion of the square does each shaded region represent?

1/4 + 1/16 + 1/64 + …

1/4

1/16

1/64

b)

The total area shaded is 1/3 of the largest square.

Independent practice

P. 63-65 #1, 5, 6, 8, 10, 12, 14, 16, 18, 19, 20,

21, 22