3. Sums of Geometric Series
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Chapter 8
3.
Sums of Geometric Series
This lesson delves into the concept of geometric series, a sequence where terms increase by a constant ratio. The focus is on understanding how to determine the sum of these series, both for finite and infinite cases. Real-life examples, such as the spread of a viral disease in a school and the behavior of a bouncing ball, are used to illustrate the concept. The lesson emphasizes the conditions under which a series converges to a specific number and when it doesn't. It also touches upon the importance of the common ratio in determining the behavior of the series. The lessons are designed to provide a comprehensive understanding, preparing students for more advanced topics.
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Lesson Settings & Tools
| | 15 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Sums of Geometric Series
Slide of 15
In the case where the terms of a sequence increase by a constant ratio, the sum of the sequence can be modeled by geometric series. This lesson will introduce how to find the sum of geometric series for both finite and infinite cases.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Finding the Sum
Consider the following sum of a sequence.
21+201+2001+20001+⋯
a Is the sum finite or infinite?
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Discussion
Geometric Series
A geometric series is the sum of the terms of a geometric sequence.
Geometric Sequence 1,2,4,8,… Geometric Series 1+2+4+8+⋯
The explicit rule of the geometric sequence above is an=2n−1. This rule can be used to write the series using sigma notation. Sum ofInfinite Termsi=1∑∞2i−1Sum ofn Termsi=1∑n2i−1
The sum of an infinite or a finite geometric series can be found by using its corresponding formula.Pop Quiz
Identifying Geometric Series
The following applet shows the first five terms of a sum. Identify whether the given sum is a geometric series or not.
Discussion
Finite Geometric Series
Let a1 and r be the first term and the common ratio, respectively, of a geometric sequence with n terms, where r=1. The sum of the related finite geometric series can be found by using the following formula.
Sn=1−ra1(1−rn), r=1
Proof
Finite SumTo derive the formula for a geometric series, a geometric sequence of n terms with the first term a1 and the common ratio r will be considered.
Notice that the like terms are ordered in minus-plus pairs. This means that after simplifying, they will cancel out and only the first and last terms will remain.
The formula for the sum of a finite geometric series has been derived.
a1,a1r,a1r2,…,a1rn−1
All the terms will be added to express the series and a1 will be factored out.
Then, both sides of the equation will be multiplied by (1−r) and the resulting equation simplified. Sn=a1(1+r+r2+⋯+rn−1)
MultEqn
LHS⋅(1−r)=RHS⋅(1−r)
Sn(1−r)=a1(1+r+r2+⋯+rn−1)(1−r)
Distr
Distribute (1−r)
Sn(1−r)=a1(1(1−r)+r(1−r)+r2(1−r)+⋯+rn−1(1−r))
Multiply
Multiply
Sn(1−r)=a1(1−r+r−r2+r2−r3+⋯+rn−1−rn)
Sn(1−r)=a1(1−r+r−r2+r2−r3+⋯+rn−2−rn−1+rn−1−rn)
AddSubTerms
Add and subtract terms
Sn(1−r)=a1(1−rn)
DivEqn
LHS/(1−r)=RHS/(1−r)
Sn=1−ra1(1−rn)
Sn=1−ra1(1−rn)
Example
The Number of Cases of Sickness at School
Tearrik goes to North High School. One day he suddenly started feeling sick at school. Later it was determined that according to school records, Tearrik's three best friends started feeling sick the next day, nine more students stayed home sick on the third day, and so on.
Within a few days, it was discovered that a new kind of viral disease was spreading rapidly through the school, with each infected student transmitting the virus to three other students each day.
This means that a common ratio exists between the number of newly-infected students each day. Therefore, their sum represents a geometric series. With this in mind, an explicit rule can be written to find the number of newly-infected students on the nth day.
If the school does not suspend classes, there will be 1093 infected students in 7 days.
a Write an explicit rule to represent the number of students who become infected on a given day.
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b Calculate the total number of infected students that there would be in 7 days if the school does not suspend classes.
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Hint
a The number of newly-infected students each day makes a geometric sequence.
b Use the formula for the sum of a geometric series.
Solution
a It is given that the number of newly-infected students is triple the number of students that were infected the day before.
an=a1⋅rn−1
Since only Tearrik fell ill on the first day, the first term of the sequence is a1=1 and the common ratio r=3. Substitute these values into the formula to find the explicit rule for this sequence.
The number of newly-infected students on the nth day can be represented by the formula an=3n−1.
b To calculate the total number of infected students on any given day, the formula for the sum of a finite geometric series will be used.
Sn=1−ra1(1−rn)
Since the total number of infected students on the 7th day is required, n=7 will be substituted into the formula. Remember that the first term a1 is 1 and the common ratio r is 3.
Sn=1−ra1(1−rn)
SubstituteValues
Substitute values
S7=1−31(1−37)
▼
Evaluate right-hand side
IdPropMult
Identity Property of Multiplication
S7=1−31−37
ExpandFrac
ba=b⋅(-1)a⋅(-1)
S7=3−137−1
CalcPow
Calculate power
S7=3−12187−1
SubTerms
Subtract terms
S7=22186
CalcQuot
Calculate quotient
S7=1093
Example
Geometric Series in Sigma Notation
As the number of infected students increased, the school decided to start a quarantine period. During the quarantine period at North High School, lessons started being taught online. In one of his math lessons, Tearrik's math teacher introduced a geometric series an example written using sigma notation.
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Hint
Use the formula for the sum of a finite geometric series.
Solution
The given sum is a finite geometric series. Therefore, the formula for the sum of a finite geometric series can be used to find this sum.
Sn=1−ra1(1−rn),r=1
In this formula, a1 is the first term, r is the common ratio, and n is the number of terms. First, substitute i=1 into the expression given on the right side of the sigma notation to find a1.
The first term is 18. Notice that the summation index i is placed as a single exponent of 3, meaning that the next term can be found by multiplying the previous one by 3. This makes the common ratio r=3. Another way to find r is to divide the second term by the first. Start by substituting i=2 into the given expression.
Now calculate the ratio of a2 to a1. a1a2=1854 ⇒ r=3
The common ratio was found to be 3. Since the lower limit of the sigma notation is 1 and the upper limit of the sigma notation is 12, there are 12 terms in the series in total, so n=12. The sum can be calculated with these values. To do so, substitute all the values into the formula for Sn and simplify.
Pop Quiz
Calculating Finite Geometric Series
Calculate the sum of all the terms of the given finite geometric series written in summation notation. Remember that the formula for the sum of a geometric series can be used rather than adding the terms one by one.
Discussion
Infinite Geometric Series
Let a1 and r be the first term and the common ratio, respectively, of a geometric sequence with n terms, where r=1. For an infinite series, if the common ratio r is greater than -1 and less than 1 — in other words, if ∣r∣<1 — then the sum can be found by using the following formula.
S∞=1−ra1, -1<r<1
This means that the sum converges on a number. If the common ratio r is less than or equal to -1 or greater than or equal to 1 — if ∣r∣≥1 — then the sum diverges. In such cases, there is no sum for the infinite geometric series.
Proof
Infinite SumTo derive the formula for the sum of an infinite geometric series with -1<r<1, the standard formula for the finite geometric series will be considered.
The formula for the sum of an infinite geometric series with -1<r<1 has been derived.
Sn=1−ra1(1−rn)
Since r is a number between -1 and 1, the value of rn becomes very small as the value of n increases. In other words, it gets closer to 0 as n approaches infinity.
rn1n→∞0
Therefore, rn=0 can be substituted into the standard formula and the resulting equation simplified. Sn=1−ra1(1−rn)
SubstituteII
rn=0, n=∞
S∞=1−ra1(1−0)
SubTerm
Subtract term
S∞=1−ra1(1)
IdPropMult
Identity Property of Multiplication
S∞=1−ra1
S∞=1−ra1
Example
Solving Examples of Geometric Series
In the next math lesson, Tearrik's math teacher continued with the topic geometric series and this time she introduced the first few terms of two different geometric series as examples.
As shown, the common ratio is 32. Since the absolute value of 32 less than 1, it means that the sum of the series does converge to a number.
The sum of the infinite geometric series given in Example I is 3.
The common ratio of the given series in Example II is -45. 
Now the absolute value of the common ratio will be found.
a Determine whether the geometric series in Example I converges or diverges. If it converges, find its sum.
b Determine whether the geometric series in Example II converges or diverges. If it converges, find its sum.
Answer
a Does the series converge or diverge? Converge
Sum: 3
b Does the series converge or diverge? Diverge
Sum: No sum.
Hint
a Consider the common ratio of the series. When does a series converge or diverge?
b What is the common ratio of the given series?
Solution
a To determine whether the sum of the series converges to a number or if the series does not have a sum, start by looking at the common ratio between the terms of the given series in Example I.
∣r∣=∣∣∣∣∣32∣∣∣∣∣⇒∣r∣=32<1
Therefore, it is possible to find its sum by using the formula for the sum of an infinite series.
S∞=1−ra1
To find the sum, substitute a1=1 and r=32 into the formula and evaluate.
S∞=1−ra1
SubstituteII
a1=1, r=32
S∞=1−321
▼
Evaluate right-hand side
OneToFrac
Rewrite 1 as 33
S∞=33−321
SubFrac
Subtract fractions
S∞=311
DivOneByFracD
a/b1=ab
S∞=13
SimpQuot
Simplify quotient
S∞=3
b Once again, start by identifying the common ratio of the given series. One way to find the common ratio is to divide the second term by the first term.
a1a2
SubstituteII
a1=-410, a2=1650
-410−1650−
▼
Simplify
DivFracByFracD
c/da/b=ba⋅cd
1650⋅(-104)
MoveNegFracToNum
Put minus sign in numerator
1650⋅(10-4)
MultFrac
Multiply fractions
160-200
ReduceFrac
ba=b/40a/40
4-5
MoveNegNumToFrac
Put minus sign in front of fraction
-45
∣r∣=∣∣∣∣∣-45∣∣∣∣∣⇒∣r∣=45>1
Since the absolute value of the common ratio is greater than 1, the series diverges. Therefore, it is not possible to find a sum for this series.
Pop Quiz
Does It Converge or Diverge?
Determine whether the given infinite geometric series converge or diverge. Remember that if the common ratio ∣r∣<1, then the infinite series converges to a number, and that if ∣r∣≥1, then the series diverges.
Discussion
Partial Sum of an Infinite Geometric Series
If the common ratio of an infinite geometric series is less than or equal to -1 or greater than or equal to 1, the sum of the series does not exist. However, it is possible to find a partial sum or the sum of the first several terms in the series. This partial series can be thought of as a finite series. As such, its sum can be found using the formula for a finite geometric series.
Sn=1−ra1(1−rn), r=1
Example
Calculating the Distance Traveled by the Ball
After recovering from his illness, Tearrik returns to school and continues to play basketball with his best friend Tadeo. Suppose that after the ball hits the rim of the basket, the ball falls 3 meters and rebounds to 85% of the height of the previous bounce.
a Find the total vertical distance traveled by the ball before it comes to rest.
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b What would the sum of the vertical distance traveled by the ball be if Tadeo caught the ball at the top of the fourth bounce? Round the answer to two decimal places.
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Hint
a The vertical distance traveled by the ball for one bounce will be 2 times 0.85 of the previous bounce since the ball rises and then falls the same distance.
b Use the formula for the partial sum of an infinite series.
Solution
a It is given that after each bounce, the ball rises 85% of the height of the previous bounce. Since the initial height of the ball is 3 meters, the first bounce will be 85%=0.85 times the initial height.
Initial Height:First Bounce:3 m3(0.85) m
For the second bounce, the height of the ball will be 0.85 times the first bounce, 3(0.85) meters.
Second Bounce:3(0.85)(0.85) ⇒ 3(0.85)2 m
The heights of the other bounces can be written by considering this pattern. 
3+2⋅3(0.85)+2⋅3(0.85)2+⋯⇓3+6(0.85)+6(0.85)2+⋯
Since a common ratio r=0.85 exists between the distances that the ball travels, the distances traveled starting with the first bounce represent a geometric series. This sum is considered infinite because it is assumed that the ball could continue to bounce in increasingly small increments forever. Now, express the sum using summation notation.
3+n=1∑∞6(0.85)n
Note that even though the sum of the series is infinite, it can still be found by using the formula for the sum of an infinite series because the absolute value of the common ratio ∣r∣=0.85 is less than 1.
S∞=1−ra1
The first term of the infinite geometric series is 6(0.85) and the common ratio is 0.85. Now, substitute these values into the formula!
n=1∑∞6(0.85)n = 1−0.856(0.85)
Then, evaluate the right hand side.
n=1∑∞6(0.85)n=1−0.856(0.85)
n=1∑∞6(0.85)n=34
34+3=37
This means that the ball will have traveled 37 meters when it comes to rest.
b This time the ball is caught at the top of the fourth bounce. To find the vertical distance traveled by the ball, start by considering the vertical distance traveled for each bounce until the end of the third bounce. Then, add the fourth vertical distance only once since the ball will rise and but be caught before it falls again.
3+2⋅3(0.85)+2⋅3(0.85)2+2⋅3(0.85)3⇓3+6(0.85)+6(0.85)2+6(0.85)3
Notice that this sum represents a partial sum of the infinite series considered in Part A. To calculate this partial sum, the formula for the sum of a finite geometric series will be used, as the sum is calculated up until n=3.
Sn=1−ra1(1−rn)
With this in mind, substitute n=3, a1=6(0.85), and r=0.85 into the formula and evaluate the result.
Now, add the fourth vertical distance 3(0.85)4 to this sum.
13.11975+3(0.85)4
CalcPow
Calculate power
13.11975+3⋅0.522006…
Multiply
Multiply
13.11975+1.566018…
AddTerms
Add terms
14.685768…
RoundDec
Round to 2 decimal place(s)
≈14.69
14.69+3=17.69
The ball will travel vertically about 17.69 meters in total until Tadeo catches it at the top of the fourth bounce.
Example
Who Will Save More Money?
Now that they are completely recovered, Tearrik and Tadeo decide to save money so they can go to the NBA finals next year, 15 months from now. They start chatting about their own ways to save money.
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Hint
Recall that a series is geometric if it has a common ratio.
Solution
To determine who will save more money in 15 months, the total amounts saved will be calculated one at a time.
Tadeo's Savings
Tadeo is planning to save 100 dollars every month. This means that his savings will increase linearly by a constant rate of 100 dollars. Therefore, multiply 100 dollars by 15 months to find the total amount of money he will have saved in 15 months.100 ⋅ 15=1500
After 15 months, Tadeo will have saved 1500 dollars. Tearrik's Savings
Tearrik will start by saving only $0.50 the first month. Then, he will increase the amount of money he saves every month by doubling the previous amount.
Sn=1−ra1(1−rn)
Now, substitute n=15, a1=0.50, and r=2 into this formula and simplify.
After 15 months, Tearrik will have saved 16383.5 dollars. Conclusion
Notice that Tearrik's savings will increase rapidly although he starts with an extremely small amount. On the other hand, Tadeo will save the same amount of money each month and his savings will therefore increase at a constant rate. Finally, after 15 months, Tearrik will have saved much more money than Tadeo.
Closure
Finding the Sum
At the beginning of the lesson, the following sum of a sequence was presented.
Therefore, the sum represents a geometric series. Notice that the absolute value of the common ratio ∣r∣=∣∣∣101∣∣∣ is less than 1.
The sum of the geometric series converges to the decimal number 0.56.
21+201+2001+20001+⋯
a Is the sum finite or infinite?
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Hint
a Check the common ratio between the terms of the sum.
b Recall the formula for the sum of an infinite series.
Solution
a As can be seen, the given sequence has infinitely many terms. That said, it can still have a finite sum. To check whether the sum converges to a number or not, the common ratio between the terms will be identified. To do so, the ratio of the second term to the first term will be calculated.
a1a2
SubstituteII
a1=21, a2=201
21 201
DivFracByFracD
c/da/b=ba⋅cd
201⋅12
MultFrac
Multiply fractions
202
ReduceFrac
ba=b/2a/2
101
∣r∣=∣∣∣∣∣101∣∣∣∣∣⇒∣r∣=101<1
This means that the sum is finite and can be calculated by using the formula for the sum of an infinite series.
b As it is mentioned in Part A, the sum is finite. Having a geometric series with a common ratio less than 1, the sum of this series will be calculated by using the formula for the sum of an infinite series.
S∞=1−ra1
Substitute a1=21 and r=101 into the formula.
S∞=1−ra1
SubstituteII
a1=21, r=101
S∞=1−10121
▼
Evaluate right-hand side
OneToFrac
Rewrite 1 as 1010
S∞=1010−10121
SubFrac
Subtract fractions
S∞=109 21
DivFracByFracD
c/da/b=ba⋅cd
S∞=21⋅910
MultFrac
Multiply fractions
S∞=1810
ReduceFrac
ba=b/2a/2
S∞=95
UseCalc
Use a calculator
S∞=0.555555…
RoundDec
Round to 2 decimal place(s)
S∞≈0.56
Sums of Geometric Series
Exercise 2.1
In a skydiving formation with n rings, each ring after the first has twice as many skydivers as the preceding ring.
A team of Argentinian skydivers is trying to set a new record. The first ring of their formation consists of five skydivers.
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Let a_n be the number of divers in the n^(th) ring. Recall that each ring after the first has twice as many skydivers as the preceding ring. Therefore, the number of skydivers in each ring form a geometric sequence. To determine its rule, we will use the formula for the explicit rule of a geometric sequence. a_n= a_1 r^(n-1) In our case, since there are 5 divers in the first ring, the first term is a_1= 5. Because each ring after the first has twice as many skydivers as the preceding ring, the common ratio is r= 2. We can substitute these values into the above rule.
We want to find the total number of skydivers when there are five rings. From Part A we know that the number of skydivers in the n^(th) ring is represented by a geometric sequence, with first term a_1= 5 and common ratio r= 2. Therefore, we can use the formula for partial sums of geometric series. S_n=a_1(1- r^n)/1- r We want to calculate the sum of the first 5 terms. So, n= 5. Let's substitute the values and find the sum.
Therefore, there are 155 skydivers in all five rings.
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A special fractal created using unit squares is called the Sierpiński Carpet. The process of creating this fractal involves removing smaller squares from larger squares and can be carried out by three steps.
1. Divide a square with side length 1 into nine congruent squares.
2. Remove the center square.
3. Repeat these steps for each smaller square.
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We want to find a rule for a_n, the total number of squares removed in the n^(th) stage. Let's analyze the steps involved in the creation of the Sierpiński Carpet.
At stage 1, we remove one square. Therefore, we know that a_1= 1. Furthermore, for each square we remove, we create 8 new squares. This means that we have a geometric sequence with common ratio r= 8. To determine a rule for this sequence, we will use the formula for the explicit rule of a geometric sequence.
To find the total number of squares removed through Stage 8, we need to find the sum of the first 8 terms of the sequence we found in Part A. To do this, we will consider the geometric series that is made by the sum of the terms of our sequence. We will use formula for the partial sum of a geometric series. S_n=a_1(1-r^n)/1-r In our case, we have that a_1= 1, r= 8, and n= 8. Let's substitute these values and find the value of the sum.
Therefore, the total number of squares removed through Stage 8 is 2 396 745.
We want to find a rule for b_n, the remaining area of the original square after the n^(th) stage. Let's analyze the steps involved in the creation of the Sierpiński Carpet again.
In Stage 1, we remove one out of the nine squares. Therefore, there are eight remaining squares in Stage 1. This means that the remaining area of the original square is 89. With this information, we can say that b_1= 89. b_1=8/9 Notice that we do the same thing in each subsequent step. We divide our figure into a congruent smaller squares, and then 8 out of 9 smaller squares remain from each square. Therefore, b_n is a geometric sequence and its common ratio is r= 89. r=8/9 To determine a rule for the geometric sequence, we will use the formula for the explicit rule of a geometric sequence.
Finally, we want to find the remaining area of the original square after Stage 12. Therefore, we need to find the value of b_(12). Let's do it!
The remaining area of the original square after Stage 12 is about 0.243 square units.
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Tearrik pushes his younger cousin Tadeo on a tire swing one time and then allows him to swing freely. On the first swing, Tadeo travels a distance of 14 feet. On each successive swing, Tadeo travels 75% of the distance of the previous swing.
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Let's analyze the distances that Tadeo travels on the tire swing.
When pushed by Tearrik, Tadeo travels 14 feet on the first swing. On each successive swing, he travels 75 % — or 0.75 — of the previous swing's distance. The total distance traveled by Tadeo is given by the infinite geometric series. 14 + 14(0.75) + 14(0.75)^2 + 14(0.75)^3 + ... For this series, the first term is a_1= 14 and the common ratio is r= 0.75. The absolute value of 0.75 is also 0.75, which is less than 1. |0.75|=0.75<1 Therefore, we can use the formula for the sum of an infinite geometric series. Let's do it!
Tadeo travels a total distance of 56 feet.
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Consider the following finite geometric series.
i=1∑k3(2)i−1=189
Find the upper limit of the summation k.
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From the given summation notation, we can see that the series is a geometric series with common ratio r=2. Let's start by finding the first term. To do this, we can substitute i= 1 in the corresponding expression.
The first term of the series is a_1=3. Let's now recall the formula for the sum of a finite geometric series. S_n=a_1(1-r^n)/1-r Here n is the upper limit of the summation, S_n is the sum of the series, and a_1 and r are the first term and the common ratio of the series, respectively. We also know that the sum of the series is 189. To find the value of k, we can substitute n= k, r= 2, a_1= 3, and S_n= 189 in the above equation, and solve for k. Let's do it!
We found that 2 raised to the power of k is the same as 2 raised to the power of 6. By the Property of Equality for Exponential Equations, this means that k=6. 2^k=2^6 ⇔ k=6 The upper limit of the given series is 6.
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Sums of Geometric Series
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