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