CLASS NOTES (COMPLETED – NO NEED TO COPY NOTES FROM
OVERHEAD) ... To establish basic elements of arithmetic sequences and series.
Example ...
1 MA40S PRE-CALCULUS UNIT G – GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED – NO NEED TO COPY NOTES FROM OVERHEAD)
Basic Elements of Arithmetic Sequences and Series Objective: • To establish basic elements of arithmetic sequences and series Example 1: Consider the arithmetic sequence 3, 7, 11, 15, 19, … What does the ‘…’ mean? What is the 7th term, t7? sequence?
What is the 101st term of the
The key feature of an arithmetic sequence is that there is a common difference d between any two consecutive terms. To obtain any term, add d to the preceding term or subtract d from the following term. We call it arithmetic series because of this adding and subtracting. The formula for the nth term ( tn ) of an arithmetic sequence is given by: a. recursively: which means that each term is calculated from the immediately preceding term tn+1 = tn + d b.
explicitly (or directly):
Which means any term can be directly calculated
tn = a + (n – 1)d where a is the first term (t1), and d is the common difference between terms. In the case of finding the 101st term then, we have a = 3, d = 4, and n = 101. t101 = 3 + (101 – 1)*4 = 403
MA40SP_G_GeoSeries_ClassNotes.doc
Revised:20110415
2 Example 2: How many terms are there in the arithmetic sequence 4, 15, 26, … , 2853? We will use the formula tn = a + (n – 1)d and will solve for the variable n. 4 + (n − 1)11 = 2853 (n − 1) =
2853 − 4 = 259 11
n = 260
Example 3: The 3rd term of an arithmetic progression is 13, and the 500th term is 2498. Find the first two terms. First, we express 13 and 2498 in terms of the formulas for t3 and t500 respectively. 13 = a + (3 - 1)d 2498 = a + (500 - 1)d
This results in a system of two equations in two unknowns: a + 2d = 13
(1)
a + 499d = 2498
(2)
Subtracting equation (1) from equation (2) we can eliminate the variable a and find d. 497d = 2485 d=5
Thus, from equation (1), we can solve for a. a = 13 − 2d a = 13 − 10 = 3 Thus the first two terms of the sequence are 3 and 8.
3 Example 4:
Find the sum of 101 terms of the series 3 + 7 + 11+ 15 + 19 + ... . The formula for the sum of n terms of an arithmetic series. n S = [2a + (n − 1)d] 2 S = 101 [ 2 × 3 + (101 - 1)4 ] 2 = 101 [ 406 ] 2 = 20503 The formula is much simpler when a =1 and d = 1. Try it for that
Teacher will prove where this formula comes from:
PRACTICE PROBLEMS YOU CAN TRY a.
find the next arithmetic term and state the ‘d’: (1) (2) (3) b.
1, 3, 5, 7, 9,…. 1, 5, 9, 13, 17, …. 22, 33, 44, 55,…
find the 12th term explicitly: (1) (2) (3)
1, 3, 5, 7, 9,…. 1, 5, 9, 13, 17, …. 22, 33, 44, 55,…
d=_____ d=_____ d=_____
4 c.
Find the sum of the arithmetic series: (1) (2) (3) (4) (5) (6)
1, 3, 5, 7, 9,…. For the first 20 terms 1, 5, 9, 13, 17, …. For the first 30 terms 22, 33, 44, 55,…For the first 10 terms what is the sum of all the numbers from 1 to 100? what is the sum of all the numbers from 1 to 1,000? what is the sum of all the numbers on a clock?
Geometric Sequences Objective: • To learn about the properties of geometric sequences Investigation:
Consider the sequence 2, 6, 18, 54. Then fill in the blanks below with t5 , t6 , and t7 . 2, 6, 18, 54, 162, 486, 1458 ↑ ↑ ↑ ↑ ↑ ↑ ↑ t1 t2 t3 t4 t5 t 6 t7 Explain the pattern whereby you found t5 , t6 , and t7 . Each term is obtained by multiplying the preceding term by 3.
This sequence of numbers is called a geometric sequence. The distinguishing feature of a geometric sequence is that the ratio formed by dividing any term (the nth term) by the preceding term (the (n–1)th term) is a constant. So above, each successive term could be calculated by tn+1 = tn*r. This is called the recursive formula for the geometric series. This ratio, r, is called the common ratio. In this example, the common ratio is 3. 1458 486 162 54 18 6 = = = = = =3 486 162 54 18 6 2
You were able to find t5 , t6 , and t7 . Can you find t21 ? Doing it ‘recursively’ means you would need to calculate all the terms t8, t9, all the way up to t20, so you could find t21.
5 There are two ways to find t21 : a longer tedious way, and a shorter more efficient way. Longer, cumbersome way: Multiply continuously by 3 until you reach t 21 . Shorter, neater way: Look for a pattern________________________ t1 2
2×3
t2 0
6
t3
2× 3
1
18
2×3
t4 2
54
2×3
t5 3
162
2× 3
t6
t7
486 5
1458 6
2× 3
4
2× 3
From this pattern, what is t21 ? t 21 = 2 × 3 = 69735668802 20
Generalization of Geometric Sequences Let the first term of the sequence be a, and let r equal the common ratio. The first seven terms can be written as follows: t1 t2 t 3 t4 t5 t6 t7 ... a ar1 ar 2 ar 3 ar4 ar5 ar 6 ... What is the formula for the 100 th term? The 610 th term? t100 = ar100−1 = ar 99 t610 = ar
610−1
= ar
609
What is the explicit formula for computing n th term of a geometric sequence directly?
t n = ar
n −1
Example 1: 1 Determine the common ratio of the sequence 1, − 13 , 19 , − 27 .
r=
−
1 3
1
=− 1 3
6 Example 2:
Determine the 14th term of the geometric sequence 6 , 12 , 24 … We have a = 6 and r = 2 Since tn = ar n −1 we have t 14 = 6 ⋅ 214 − 1 = 6 ⋅ 2 13 = 49152
Notice that geometric series are very much just exponential equations (from that unit!), the variable is an exponent. In this case though, the variable can only take on whole numbers Example 3:
Using an r of less than 1 (r < 1)
In the World Dominoes tournament 78,125 players are placed in groups of 5 players per table. One game is played by these 5 players, and the winner at each table advances to the next round, and so on until the final game of 5 players. How many rounds would the ultimate winner have played (including the final round)? We have a = 78125; t n = 5; r = 1 . 5 t n = ar n − 1 5 = 78125 1 5
n −1
Finally, n = 7. The winner would have played 7 rounds.
Example 5:
Relating to exponential equations
The world’s population grows by 2% per year. The world food production can sustain an additional 200 million people per year. In 1987 the population was 5 billion, and food production could sustain 6 billion people. a) Calculate the population in 1998, 2009, and 2019. b) Calculate the population that food production could sustain in 1998, 2009, 2019. c) When will the population exceed the food supply?
7 a) The world’s population in each of the years starting with 1987 can be considered as a geometric sequence. The first term of the sequence will be 5 (billion) and the common ratio r will be 1.02. Fill in the table below with the world population for the first five years, starting with 1987. t1 1987 5 billion
t2 1988 5 × (1.02)1 = 5.1 billion
t3 1989 5 × (1.02)2 = 5.2 billion
t4 1990 5 × (1.02)3 = 5.3 billion
t5 1991 5 × (1.02)4 = 5.4 billion
n −1 So the formula for the world population in any year is P = 5 × (1.02)
where n is the
position of that year in the sequence of years starting with 1987. What is the value of n for each of the following years ? 1998: n =
12
2009: n =
23
2019: n =
33
Now calculate the world population in each of these years. 1998 2009
2019
P = 5 × (1.02)11
P = 5 × (1.02)22
P = 5 × (1.02)32
= 6.2 billion
= 7.7 billion
= 9.4 billion
b) The world’s population that food production can sustain in each of the years starting with 1987 can be considered as an arithmetic sequence. The first term t1 = a of the sequence will be 6 (billion) and the common difference d will be 0.2 (billion). Fill in the table below with the world sustainable population for the first five years, starting with 1987. t1 1987 6 billion
t2 1988 6 + (2 − 1)0.2 = 6.2 billion
t3 1989 6 + (3 − 1)0.2 = 6.4 billion
t4 1990 6 + (4 − 1)0.2 = 6.6 billion
t5 1991 6 + (5 − 1)0.2 = 6.8 billion
8
So the formula for the world population in any year is P = 6 + (n − 1)0.2
where n is the
position of that year in the sequence of years starting with 1987. Like before, fill in the value of n for each of the following years. 1998: n =
12
2009: n =
23
2019: n =
33
Now calculate the world sustainable population in each of these years. 1998 P = 6 + (12 − 1)0.2 = 8.2 billion
2009 P = 6 + (23 − 1)0.2 = 10.4 billion
2019 P = 6 + (33 − 1)0.2 = 12.4 billion
c) To find when the world population will exceed the sustainable population, we equate the two formulas and solve for n.
5 × (1.02)
n −1
= 6 + (n − 1)(0.2)
We’ll use the graphing calculator to solve this rather complex equation. Sketch the intersecting curves in the blank grid below.
Using the intersect tool, we find that n equals approximately __71__ years. Therefore, the world population will exceed sustainable population in about the year 1987 + (71–1) = 2057 when the world population is about
9
billion.
20 YOU TRY: a.
Find the 8th term of the sequence 3,6,12, 24,…
b. A sum of money is invested and compounds annually. Find the value of $1,000 invested at 6% after 10 years.
Geometric Series A ‘series’ is just the sum of the terms of a ‘sequence’ Objective: • To derive and apply expressions representing sums for geometric growth and to solve problems involving geometric series Definition: A geometric series is the
sum
of the terms of a geometric sequence
Consider the geometric sequence 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 . If we add the terms of the sequence, we can write the geometric series as Sum = S = 1 + 2 + 4 + 8 +16 + 32 +64 +128 + 256 +512 +1024 . Instead of adding the terms directly, let’s evaluate S using a trick as follows. We will multiply each term of the series by 2 (which is the value of the common ratio r), realign the result under the original series, and subtract equation (2) from equation (1). S = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 ....................(1) 2S = 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 .........(2) S − 2S = 1− 2048 − S = −2047 S = 2047 The above is not unlike the elimination method we used previously in Systems of Linear Equations.
.
10 We can use this process to develop a general formula for Sn , the sum of the first n terms of a geometric series. In general, a geometric series can be written as follows, with rSn written underneath. Sn = a + ar + ar 2 + ar 3 + ... + ar n -2 + ar n -1 ..................(1) rSn = ar + ar 2 + ar 3 + ... + ar n -2 + ar n -1 + ar n .........(2) Sn − rSn = a − ar n Factoring Sn , we have Sn (1− r) = a − ar n Finally, Sn =
a − ar n a(1− r n ) = 1− r 1− r where r ≠ 1
Alternatively, since tn = ar n-1 we can write Sn as follows: n
Sn =
a( 1− r ) (1− r )
a − ar n a − ar n -1 r a − rt n a − rl = = = 1− r 1− r 1− r 1− r where l is the last term of the geometric series (i.e., t n or ar n −1 ) =
Summary: The sum of the first n terms of a geometric series is given by: a(1− r n ) Sn = 1− r or where a is the first term, l is the last term, and r ≠ 1. a − lr Sn = 1− r
Example 1: Determine the sum of 14 terms of the geometric series: S = 6 + 18 + 54 + ...
a(1 − r n ) 1− r 6(1− 314 ) = = 14348904 1− 3 Sn =
11
Example 2: Determine the n th term, and the sum of the first n terms of the geometric sequence which has 2, 6, and 18 as its first three numbers. t n = ar n-1 t n = 2(3)n-1
Sn =
a(1 - r n ) (1 - r)
Sn =
2(1 - 3 n ) = 3n - 1 (1 - 3)
Example 3: For the geometric series 1 − 1 + 1 − 1 + ..., find the sum of the first 20 terms. 3 9 27 a(1 − r n ) Sn = (1 − r)
1 20 11− − 20 3 3 1 ⋅ 3 = 1− = S20 = 1 4 3 4 1− − 3
Example 4:
While training for a race, a runner increases her distance by 10% each day. If she runs 2 km on the first day, what will be her total distance for 26 days of training? (accurate to 2 decimal places) a=2 r = 1.1 S26 = 2 + 2(1.1) + 2(1.1)2 + 2(1.1) 3 + ... + 2(1.1)24 + 2(1.1) 25 n = 26 26 a(1− r n ) 2(1− 1.1 ) S26 = = = 218.36 km 1− r 1 Ğ1.1
12
Sigma Notation Objective: • To use sigma notation to write and evaluate series A series can be written using the Greek capital letter ∑ (‘sigma’). This notation provides us with a mathematical “shorthand” for writing and working with various kinds of series. A series written with sigma notation could have the following form. 7
∑n
2
n =1
This notation means simply that we replace the variable n with the numbers 1 through 7 respectively, and then add the resulting terms. 7
So,
∑n
2
= 12 + 22 + 32 + 4 2 + 52 + 62 + 7 2 = 140
n =1
Part I: Converting sigma notation to expanded form Example 1: 5
Write the series
∑ 2 k in expanded form, and compute the sum. k =1
We replace the variable k with the numbers 1 through 5 respectively, and then add the resulting terms. 5
∑ 2k = 2 + 2 1
2
+ 23 + 24 + 25 = 62
k= 1
Example 2: 8
Write the series
∑(4n − 5) in expanded form, and compute the sum. n =3
We replace the variable n with the numbers 3 through 8 respectively, and then add the resulting terms. Note that the bottom “index” does not necessarily have to equal 1.
13 8
∑(4n − 5) = (4 ⋅ 3 − 5) + (4 ⋅ 4 − 5) + (4 ⋅ 5 − 5) + (4 ⋅6 − 5) + (4 ⋅7 − 5) + (4 ⋅ 8 − 5) n=3
= 7 + 11+ 15 + 19 + 23 + 27 = 62 (7 + 27) = 102 Note that this is an arithmetic series, just an arithmetic sequence with the terms added together. Example 3: 4
Write the series
∑(−1)
i− 1
4 i in expanded form, and compute the sum.
i =1
As before, we replace the variable i with the numbers 1 through 4 inclusive, and then add the resulting terms. 4
∑(−1)
i −1
1− 1
4i = (−1)
2− 1
⋅ 4( 1) + (−1)
3 −1
⋅ 4( 2) + (−1)
⋅ 4 (3) + (−1)
4− 1
⋅ 4(4 )
i=1
= (−1) 4 (1) + (−1) 4(2) + (−1) 4 (3) + (−1) 4( 4) 0
= 4 = −8
1
+
(−8)
2
+
12
3
+
(−16 )
Part II: Converting expanded form to sigma notation Example 4:
Convert the series 5 + 9 + 13 + 17 + 21 + 25 + 29 to sigma notation. There are two steps in the conversion process. Step 1 Find a formula for the general term tn .
Since this is an arithmetic series, we can compute t n as follows. t n = a + (n − 1)d = 5 + (n − 1)4 = 4n + 1
14 Step 2 Since there are 7 terms, we place n = 1 under the letter sigma, and the index 7 above the letter sigma. Then we write the expression (4 n − 1) to the right of the symbol. 7
∑ (4n − 1) n=1
Example 5:
Convert the series 3 + 12 + 48 + 192 + 768 + 3072 to sigma notation. Step 1 Find a formula for the general term tn .
Since this is a geometric series, we can compute t n as follows. t n = ar n − 1 tn = 3 ⋅ 4
n −1
Step 2 Since there are 6 terms, we place n = 1 under the letter sigma, and the index 6 above the letter sigma. Then we write the expression 3 ⋅ 4 n−1 to the right of the symbol 6
∑ 3⋅ 4
n−1
n=1
Example 6: 50
How many terms are there in the series
∑3
k +2
?
k =1
50 terms
Example 7: 9
How many terms are there in the series
∑3
k +2
?
k =4
You might give the answer 9 − 4 = 5 terms, but you can see by expanding the series that the answer is actually 6 terms. That is because we start counting (zero), at the 4th term which is included)
15 Example 8: 800
How many terms are there in the series
∑3
k +2
?
k =4
Rule: Subtract the bottom index from the top index, and add 1. The number of terms is (800 − 4) + 1 = 797 You Try some problems
1.
Write each of these in expanded notation and then calculate their sums 5
a.
∑ 2k k =1
4
b.
∑k
2
k =1
6
c.
∑ ( 2 k − 4) k =1
4
d.
∑2
k
k =1
2.
3.
4.
Write the following series using Sigma notation a.
3 + 9 + 27 + 81
b.
6 + 8 + 10 + 12 + 14 + 16 + 18 + 20
c.
–1 + 4 – 9 + 16 – 25 + 36
Consider the sequence 3, 6, 12, 24,…. a.
find the 8th term
b.
find the sum of the first 8 terms
Consider the geometric sequence with t1 = 1000 and r = 1.05 a.
write the first three terms
b.
find the sum of the first 20 terms (to two decimal places)
16
5.
Evaluate: 17
a.
∑ 8(1.2) k
5
b.
k =3
15
∑2 k =1
k
Infinite Geometric Series Objective: • To investigate the concept of infinite geometric series Investigation: 1 1 1 ... Consider the geometric series S = 12 + 14 + 18 + 16 + + + 32 64 Now compute the following partial sums for the series.
3 S2 = 1 + 1 = 2 4 4 7 S3 = 12 + 1 + 1 = 4 8 8 15 S4 = 12 + 1 + 1 + 1 = 4 8 16 16 1 + 1 = 31 S5 = 12 + 14 + 18 + 16 32 32
63 S6 = 12 + 1 + 1 + 1 + 1 + 1 = 4 8 16 32 64 64 1 + 1 + 1 + 1 = 127 S7 = 12 + 14 + 18 + 16 32 64 128 128 1 = 255 S8 = 12 + 1 + 1 + 1 + 1 + 1 + 1 + 256 4 8 16 32 64 128 256
511 S9 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 2 4 8 16 32 64 128 256 512 512 1 + 1 + 1 = 1023 S10 = 12 + 1 + 1 + 1 + 1 + 1 + 1 + 256 512 1024 4 8 16 32 64 128 1024 What do you notice about the sums for Sn as n gets larger and larger?
17
The values get closer and closer to 1.
When we increase the value of n even more, what happens to the values for Sn ?
The sums would get even closer to 1. To further investigate this phenomenon, let’s look at the algebraic expression for Sn for this particular geometric series. We have a = 12 and r = 12 .
a(1 − r n ) Sn = 1− r n 1 1− 1 2 2 = 1− 1 2 = 1− 1 2
n
n
As n gets larger and larger, what happens to 1 ? 2 It gets closer and closer to 0. n
As n grows larger and larger, how close to zero will 1 get? 2 As close as you want n
Can 1 ever equal 0? 2
No n
In other words, by making n sufficiently large, you can make 1 as close to 0 as you like. 2 This is the foundation of the mathematical concept called the limit. (You will learn in depth about limits when you study calculus.)
18 n
Formally, we say that as n increases without bound, the number 1 approaches 0, and 2 n therefore, the value of Sn = 1 − 1 approaches 1. If you were graphing it you see an 2 asymptote. n
How close can Sn = 1 − 1 get to 1? 2 By making n sufficiently large, Sn can get as close as you want to 1. 1 1 1 ... th If the series S = 12 + 14 + 18 + 16 + + + does not end at the n term, but rather continues 32 64 on indefinitely, we call it an infinite series. 1 1 1 ... + + + is __1___. We also say that the sum of the infinite series S = 12 + 14 + 18 + 16 32 64
Generalization:
Let’s look at the concept of the sum of an infinite geometric series using the formula a(1 − r n ) Sn = 1 − r ( r ≠ 1) . If r < 1, what happens to rn as n increases without bound? n r gets closer and closer to 0 How close to 0 can rn get?
By making n sufficiently large, as close as you want Thus if r < 1, we say that the sum of an infinite series is S∞ =
a(1 − 0) a = 1− r 1− r
If r > 1, what happens to the value of Sn as n gets larger and larger?
Sn also gets larger and larger. In other words, it does not have a finite sum. e.g., Sn = 1 + 2 + 4 + 8 + 16 + 32 + ... We say that the sequence associated with this series ‘diverges’, it is a divergent sequence
19 Example 1:
Find the sum of the infinite series 2 +
2 2 + ... . 5 25
a = 2 and r = So S∞ =
1 5
a 2 5 = = 1− r 1− 1 2 5
Example 2: ∞
Evaluate:
∑ 31
k
k =1
1 1 1 ... S∞ = 31 + 1 9 + 27 + 81 + 243 + We have a = 31 and r = 31 1 a so S = 1 − r = 3 1 = 1 1− 3 2 Example 3:
Find the sum of the infinite geometric series 3 3 3 3 3 3 S∞ = + + + + + + ... . 10 100 1000 10000 100000 1000000 3 1 We have a = 10 and r = . 10 3 3 a 10 10 1 So, S∞ = = = 9 =3 1− r 1− 1 10 10 3 3 3 3 3 3 Note that the series S∞ = 10 + + + + + + ... is the 100 1000 10000 100000 1000000 expanded form of the repeating decimal 0.3 .
20 Example 4:
An oil well produces 25 000 barrels of oil during its first month of production. If its production drops by 5% each month, estimate the total production before the well runs dry. S∞ = 25 000 + 25 000 ⋅ (0.95) + 25 000 ⋅ (0.95) + 25 000 ⋅ (0.95) + 25 000 ⋅ (0.95) + ... We have a = 25 000 and r = 0.95 . 25000 So, S = = 500 000 barrels. 1 − 0.95 2
3
4
Example 5:
A ball is dropped from a height of 16 metres. The ball rebounds a half of the height after each bounce. Calculate the total vertical distance the ball travels before coming to rest.
We ignore the first 16 m in computing the sum of the infinite series. S∞ = 2⋅ 8 + 2⋅ 4 + 2 ⋅ 2 + 2⋅ 1 + 2⋅ 1 + .... 2 = 16 + 8 + 4 + 2 + 1 + ... We have a = 16 and a = 1 . 2 So, S∞ = 1 −a r = 161 = 32 metres. 1− 2 The total vertical distance travelled by the ball is 32 + 16 = 48 metres.
21 SOME FOR YOU TO TRY 1.
2.
Consider the series 4 + 2 + 1 +
1 + ... 2
a.
find the 6th term
b.
find the sum of the six terms
c.
find, to 4 decimal places, the sum of the first six terms
d.
find the sum to infinity
Find the sum of each of these infinite geometric series: a.
8 8 8 + + + ... 3 9
b.
6−3+
c.
1−
3 3 − + ... 2 4
1 1 1 + − + .... 2 4 8
3. Here is a good one relating to fractals! A 4 x 4 unit square is divided into four equal squares. The bottom left square is shaded. The top right square is divided into 4 equal squares and its bottom left is shaded, then etc. This is done forever. a.
what is the total area shaded?
b. what is the perimeter of all the shaded squares?
22
4.
Investigate this series (it is of course not arithmetic nor geometric):
f ( x) = 1 −
x 2 x 4 x 6 x8 + − + − + − +.... 2! 4! 6! 8!
Try plugging in 2 and 5 (ie: find f(2) and f(5)).
Compare that with cosine of 2 and cosine of 5 (in radians of course). Try graphing f(x) and cos(x). Compare!
