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| Student Learning Objectives: |
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| | 15 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Consider the following sum of a sequence. 1/2 + 1/20 + 1/200 + 1/2000 + ⋯
A geometric series is the sum of the terms of a geometric sequence. c|c Geometric & Geometric Sequence & Series [0.5em] 1,2,4,8,... & 1+2+4+8+⋯ The explicit rule of the geometric sequence above is a_n=2^(n-1). This rule can be used to write the series using sigma notation. ccc Sum of & & Sum of Infinite Terms & & nTerms [0.5em] ∑ _(i=1)^(∞) 2^(i-1) & & ∑ _(i=1)^n 2^(i-1)
The sum of an infinite or a finite geometric series can be found by using its corresponding formula.The following applet shows the first five terms of a sum. Identify whether the given sum is a geometric series or not.
Let a_1 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.
S_n=a_1(1-r^n)/1-r, r≠ 1
Then, both sides of the equation will be multiplied by (1-r) and the resulting equation simplified.
LHS * (1-r)=RHS* (1-r)
Distribute (1-r)
Multiply
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.
Add and subtract terms
.LHS /(1-r).=.RHS /(1-r).
The formula for the sum of a finite geometric series has been derived.
S_n=a_1(1-r^n)/1-r
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.
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 n^\text{th} day. a_n=a_1 * r^(n-1) Since only Tearrik fell ill on the first day, the first term of the sequence is a_1= 1 and the common ratio r= 3. Substitute these values into the formula to find the explicit rule for this sequence.
a_1= 1, r= 3
Identity Property of Multiplication
The number of newly-infected students on the n^\text{th} day can be represented by the formula a_n=3^(n-1).
S_n=a_1(1-r^n)/1-r Since the total number of infected students on the 7^\text{th} day is required, n= 7 will be substituted into the formula. Remember that the first term a_1 is 1 and the common ratio r is 3.
Substitute values
Identity Property of Multiplication
a/b=a * (-1)/b * (-1)
Calculate power
Subtract terms
Calculate quotient
If the school does not suspend classes, there will be 1093 infected students in 7 days.
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.
Calculate the given sum.
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 a_2 to a_1. a_2/a_1=54/18 ⇒ 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 S_n and simplify.
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.
Let a_1 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_(∞)=a_1/1-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.
r^n= 0, n= ∞
Subtract term
Identity Property of Multiplication
The formula for the sum of an infinite geometric series with -1
S_(∞)=a_1/1-r
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.
Sum: 3
Sum: No sum.
As shown, the common ratio is 23. Since the absolute value of 23 less than 1, it means that the sum of the series does converge to a number. |r|=|2/3| ⇒ |r|=2/3 < 1 Therefore, it is possible to find its sum by using the formula for the sum of an infinite series. S_(∞)=a_1/1-r To find the sum, substitute a_1= 1 and r= 23 into the formula and evaluate.
a_1= 1, r= 2/3
Rewrite 1 as 3/3
Subtract fractions
1/a/b= b/a
Simplify quotient
The sum of the infinite geometric series given in Example I is 3.
a_1= - 104, a_2= 5016
.a /b./.c /d.=a/b*d/c
Put minus sign in numerator
Multiply fractions
a/b=.a /40./.b /40.
Put minus sign in front of fraction
The common ratio of the given series in Example II is - 54.
Now the absolute value of the common ratio will be found. |r|=|-5/4| ⇒ |r|=5/4 > 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.
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.
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.
S_n=a_1(1-r^n)/1-r, r≠ 1
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.
Initial Height:& 3 m [0.5em] First Bounce:& 3 (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.
Because the ball rises and then falls the same distance after each bounce, the vertical distance traveled is 2 times 0.85 of the previous bounce. Since the initial height of the ball is 3 meters, the total distance traveled by the ball can be written as follows. 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_(∞)=a_1/1- r 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 = 6(0.85)/1- 0.85 Then, evaluate the right hand side.
The sum of the series is 34. Now the initial 3-meter height of the ball will be added. 34 + 3 = 37 This means that the ball will have traveled 37 meters when it comes to rest.
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. S_n=a_1(1- r^n)/1- r With this in mind, substitute n=3, a_1=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.
Calculate power
Multiply
Add terms
Round to 2 decimal place(s)
Finally, the initial 3-meter height will be added. 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.
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.
Who will save more money in 15 months?
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 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.
Since there is a common ratio between the amount of money saved each month, this sum represents a geometric series. Therefore, the formula for the sum of a finite geometric series will be used to calculate the total amount of money Tearrik will have saved in 15 months. S_n=a_1(1- r^n)/1- r Now, substitute n= 15, a_1= 0.50, and r= 2 into this formula and simplify.
After 15 months, Tearrik will have saved 16 383.5 dollars.
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.
At the beginning of the lesson, the following sum of a sequence was presented. 1/2+1/20+1/200+1/2000+ ⋯
a_1= 1/2, a_2= 1/20
.a /b./.c /d.=a/b*d/c
Multiply fractions
a/b=.a /2./.b /2.
Notice that this ratio exists between each pair of consecutive terms.
Therefore, the sum represents a geometric series. Notice that the absolute value of the common ratio | r|=| 110| is less than 1. | r|=| 1/10| ⇒ |r|=1/10 < 1 This means that the sum is finite and can be calculated by using the formula for the sum of an infinite series.
S_(∞)=a_1/1- r Substitute a_1= 12 and r= 110 into the formula.
a_1= 12, r= 110
Rewrite 1 as 10/10
Subtract fractions
.a /b./.c /d.=a/b*d/c
Multiply fractions
a/b=.a /2./.b /2.
Use a calculator
Round to 2 decimal place(s)
The sum of the geometric series converges to the decimal number 0.56.
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.
2. Remove the center square.
3. Repeat these steps for each smaller square.
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.
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.
What is the total distance that Tadeo swings?
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.
Consider the following finite geometric series. ∑_(i=1)^k 3(2)^(i-1)=189 Find the upper limit of the summation k.
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.