bell ringer · 2019-10-17 · explicit versus recursive explicit –infinitely many sequences exist...

Bell Ringer

Sequences and SeriesChapter 10, Section 1 (McGraw Precal)

Pages 590 – 594

Lesson Objective4.1A, 4.1B

The students will investigate several different types of sequences. Students will use Sigma Notation to represent and calculate sums of series.

Demonstration of Learning

Given 5 problems involving series, the students will correctly solve at least 3 of them.

AP Calculus BC Learning Objectives

LO Students will be able to

4.1A Determine whether a series converges or diverges.

4.1B Determine or estimate the sum of a series.

REVIEW of Pre CALCULUSSequences

Series

SequenceA sequence is a list of things (usually numbers) that arein order.

3, 5, 7, 9, …

2nd term

1st term 3rd term

4th termThree dots means it goes on forever

When the sequence goes on forever it is called an INFINITE sequence, otherwise it is a FINITE sequence

{1, 2, 3, 4, … }

{20, 25, 30, 35, … }

{1, 3, 5, 7}

{4, 3, 2, 1}

Let’s look at some sequences

What do you

notice?

Example 1aFind the next four terms of the sequence

{2, 7, 12, 17, …}

Since the nth term is not given, we must look

for a possible pattern

We can look at the difference between

each term

One possible pattern is that each term is 5 greater than the previous term

Example 1aFind the next four terms of the sequence

{2, 7, 12, 17, …}

5𝑡ℎ 𝑡𝑒𝑟𝑚 = 17 + 5

6𝑡ℎ 𝑡𝑒𝑟𝑚 = 17 + 10

7𝑡ℎ 𝑡𝑒𝑟𝑚 = 17 + 15

8𝑡ℎ 𝑡𝑒𝑟𝑚 = 17 + 20

{2, 7, 12, 17, 22, 27, 32, 37,… }

= 22

= 27

= 32

= 37

Example 1bFind the next four terms of the sequence

{2, 5, 10, 17, …}

Since the nth term is not given, we must look

for a possible pattern

We can look at the sum of the difference between terms.

One possible pattern is that each sum of the difference is equal to the odd sequence {3, 5, 7, …}

Example 1bFind the next four terms of the sequence

{2, 5, 10, 17, …}

𝑎2 − 𝑎1 = 5 − 2

𝑎3 − 𝑎2 = 10 − 5

𝑎4 − 𝑎3 = 17 − 10

{3, 5, 7, … }

= 3

= 5

= 7

This new sequence is our pattern

between terms We can

assume the next 4 in the

sequence

9,11,13,15

Example 1aFind the next four terms of the sequence

{2, 5, 10, 17, …}

{3, 5, 7,9,11,13,15… }

5𝑡ℎ 𝑡𝑒𝑟𝑚 = 17 + 9

6𝑡ℎ 𝑡𝑒𝑟𝑚 = 26 + 11

7𝑡ℎ 𝑡𝑒𝑟𝑚 = 37 + 13

8𝑡ℎ 𝑡𝑒𝑟𝑚 = 50 + 15

= 𝟐𝟔

= 𝟑𝟕

= 𝟓𝟎

= 𝟔𝟓

{2, 7, 12, 17, 26, 37, 50, 65,… }

Example 1cFind the first four terms of the sequence

𝑎𝑛 = 2𝑛 −1𝑛

Since the nth term is given, we can just plug and chug!!

Example 1cFind the first four terms of the sequence

𝑎𝑛 = 2𝑛 −1𝑛

𝑎1 = 2 1 −11

𝑎2 = 2 2 −12

𝑎3 = 2 3 −13

{−2, 4, −6, 8… }

= −2

= 4

= −6

𝑎4 = 2 4 −14 = 8

𝑛 = 1

𝑛 = 2

𝑛 = 3

𝑛 = 4

Explicit versus Recursive

Explicit – Infinitely many sequences exist with the

same first few terms. To sufficiently define a

unique sequence, a formula for the nth term or

other information must be given. An explicit

formula gives the nth term an as a function of n.

Recursive – Recursively defined sequences give

one or more of the first few terms and then define

the terms that follow using those previous terms.

Example 2a

Find the fifth term of the recursively defined

sequence 𝑎1 = 2, 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 where, 𝑛 ≥ 2

Since the sequence is defined recursively, all the terms before the fifth term must be found first.

Example 2a

Find the fifth term of the recursively defined

sequence 𝑎1 = 2, 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 where, 𝑛 ≥ 2

𝑎2 = 𝑎2−1 + 2 2 − 1 𝑎2 = 5𝑛 = 2= 𝑎1 + 3

= 2 + 3

𝑎3 = 𝑎3−1 + 2 3 − 1 𝑎3 = 10𝑛 = 3

= 𝑎2 + 5

= 5 + 5𝑎4 = 𝑎4−1 + 2 4 − 1 𝑎4 = 17𝑛 = 4

= 𝑎3 + 7

= 10 + 7

Example 2a

Find the fifth term of the recursively defined

sequence 𝑎1 = 2, 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 where, 𝑛 ≥ 2

𝑎5 = 𝑎5−1 + 2 5 − 1 𝑎5 = 26𝑛 = 5= 𝑎4 + 9

= 17 + 9

Convergent versus Divergent

Convergent – A sequence whose limit approaches

a unique number.

Divergent – A sequence whose limit DOES NOT

approaches a unique number.

Example 3a

Determine whether the sequence is convergent or

divergent

𝑎𝑛 = −3𝑛 + 12

𝑎0 = −3 0 + 12 𝑎0 = 12𝑛 = 0𝑎1 = −3 1 + 12 𝑎1 = 9𝑛 = 1𝑎2 = −3 2 + 12 𝑎2 = 6𝑛 = 2

𝑎3 = −3 3 + 12 𝑎3 = 3𝑛 = 3𝑎4 = −3 4 + 12 𝑎4 = 0𝑛 = 4𝑎5 = −3 5 + 12 𝑎5 = −3𝑛 = 5𝑎6 = −3 6 + 12 𝑎6 = −6𝑛 = 6𝑎7 = −3 7 + 12 𝑎7 = −9𝑛 = 7

Example 3a

Determine whether the sequence is convergent or

divergent

𝑎𝑛 = −3𝑛 + 12

𝑎0 = 12𝑎1 = 9𝑎2 = 6

𝑎3 = 3𝑎4 = 0𝑎5 = −3𝑎6 = −6𝑎7 = −9

2 4 6 8 10 n

4

8

-4

-8

an

Example 3a

Determine whether the sequence is convergent or

divergent

𝑎𝑛 = −3𝑛 + 12

2 4 6 8 10 n

4

8

-4

-8

an

an does NOT

approach a finite

number, so it must be divergent

Example 3b

Determine whether the sequence is convergent or

divergent

𝑎1 = 36𝑛 = 1𝑎2 = −

1

236 𝑎2 = −18𝑛 = 2

𝑎3 = −1

2−18 𝑎3 = 9𝑛 = 3

𝑎4 = −1

29 𝑎4 = −4.5𝑛 = 4

𝑎5 = −1

2−4.5 𝑎5 = 2.25𝑛 = 5

𝑎6 = −1

22.25 𝑎6 = −1.125𝑛 = 6

𝑎7 = −1

2−1.125 𝑎7 = 0.5625𝑛 = 7

𝑎8 = −1

20.5625 𝑎8 = −0.2813𝑛 = 8

𝑎𝑛 = 36, 𝑎𝑛 = −1

2𝑎𝑛−1, 𝑎 ≥ 2

Example 3b

Determine whether the sequence is convergent or

divergent

2 4 6 8 10 n

24

36

-12

an

𝑎1 = 36𝑎2 = −18𝑎3 = 9𝑎4 = −4.5

𝑎5 = 2.25

𝑎6 = −1.125

𝑎7 = 0.5625

𝑎8 = −0.2813

12

𝑎𝑛 = 36, 𝑎𝑛 = −1

2𝑎𝑛−1, 𝑎 ≥ 2

Example 3b

Determine whether the sequence is convergent or

divergent

2 4 6 8 10 n

24

36

-12

an

12

𝑎𝑛 = 36, 𝑎𝑛 = −1

2𝑎𝑛−1, 𝑎 ≥ 2

an does approach a

finite number, so it must be

convergent𝐥𝐢𝐦𝒏→∞𝒂𝒏 = 𝟎

an does approach a

finite number, so it must be

convergent

Example 3c

Determine whether the sequence is convergent or

divergent

𝑎1 = −0.2 𝑎2 = 0.222

𝑎3 = −0.231 𝑎4 = 0.235

𝑎5 = −0.238 𝑎6 = 0.24

𝑎7 = −0.241 𝑎8 = 0.242

𝑎𝑛 =−1 𝑛 ∙ 𝑛

4𝑛 + 1

𝑎9 = −0.243 𝑎10 = 0.244

𝑎11 = −0.244 𝑎12 = 0.245

What pattern do

you notice?

Example 3c

Determine whether the sequence is convergent or

divergent

2 4 6 8 10 n

0.25

an

𝑎𝑛 =−1 𝑛 ∙ 𝑛

4𝑛 + 1

-0.25

an

approaches 0.25 when n is even, and -0.25 when n is

odd.

an does NOT

approach a unique finite

number, so it must be divergent

SeriesA series is the indicated sum of all the terms of asequence.

{𝟑, 𝟓, 𝟕, 𝟗, … } 𝟑 + 𝟓 + 𝟕 + 𝟗 +⋯=

When the series is a sum of an infinite sequence, we call it an INFINITE series; when it is the sum of a finite

sequence, we call it a FINITE series.

Example 4aFind the fourth Partial Sum of

We must find the first four terms.

Then find the sum of the first four terms.

𝑎𝑛 = −2𝑛 + 3

nth Partial SumAn nth partial sum is the sum of the first n terms and isdenoted by 𝑆𝑛

Example 4aFind the fourth Partial Sum of

𝑎𝑛 = −2𝑛 + 3

𝑎1 = (−2)1 + 3 𝑎1 = 1𝑛 = 1

𝑎2 = 7𝑛 = 2𝑎3 = −5𝑛 = 3𝑎4 = 19𝑛 = 4

𝑎2 = (−2)2 + 3

𝑎3 = (−2)3 + 3

𝑎4 = (−2)4 + 3

𝑆4 = 1 + 7 + −5 + 19

𝑆4 = 22

Example 4b

Find 𝑆3 of 𝑎𝑛 =4

10𝑛

𝑎1 =4

101𝑎1 = 0.4𝑛 = 1

𝑎2 = 0.04𝑛 = 2

𝑎3 = 0.004𝑛 = 3

𝑎2 =4

102

𝑎3 =4

103

𝑆3 = 0.4 + 0.04 + 0.004

𝑆3 = 0.444

Example 5aFind the sum of

Sigma NotationTo denote a series, we can use this notation:

𝑛=1

𝑘

𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 +⋯+ 𝑎𝑘

Where n is the index of summation, k is the upperbound of summation, and 1 is the lower bound ofsummation.

𝑛=1

5

4𝑛 − 3

Example 5aFind the sum of

𝑛=1

5

4𝑛 − 3

4(1) − 3 = 1𝑛 = 1= 5𝑛 = 2= 9𝑛 = 3= 13𝑛 = 4

4(2) − 34(3) − 34(4) − 3

𝑛=1

5

4𝑛 − 3 = 1 + 5 + 9 + 13 + 17

= 17𝑛 = 5 4(5) − 3

= 45

Example 5bFind the sum of

𝑛=3

76n − 3

2

6 3 − 3

2= 7.5𝑛 = 3

= 10.5𝑛 = 4

= 13.5𝑛 = 5

= 16.5𝑛 = 76 4 − 3

2

6 7 − 3

2

6 5 − 3

2

𝑛=3

76n − 3

2= 7.5 + 10.5 + 13.5 + 16.5 + 19.5

= 19.5

𝑛 = 6 6 6 − 3

2

= 67.5

Example 5cFind the sum of

𝑛=1

∞7

10n

7

101= 0.7𝑛 = 1

= 0.07𝑛 = 2

= 0.007𝑛 = 3

= 0.0007𝑛 = 57

1027

105

7

103

𝑛=1

∞7

10n= 0.7 + 0.07 + 0.007 + 0.0007 + 0.00007 +⋯

= 0.00007

𝑛 = 4 7

104

= 0.77777… . 𝑜𝑟7

9