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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.
To determine whether the given series is a geometric series, we will calculate the ratios between consecutive terms.
The ratio between consecutive terms is the same. Therefore, the series is geometric.
As we found in Part A, the ratios between consecutive terms are all equal to 12. Therefore, the common ratio of the given geometric series is 12.
r=1/2
To determine whether the given series is a geometric series, we will calculate the ratios between consecutive terms.
The ratio between consecutive terms is the same. Therefore, the series is geometric.
As we found in Part A, the ratios between consecutive terms are all equal to 13. Therefore, the common ratio of the given geometric series is 13.
r=1/3
Consider the following geometric series. ∑_(i=1)^9 6(7)^(i-1) Find the sum of this geometric series.
We want to evaluate the sum of the given geometric series. To do so, we will use the formula for the sum of a geometric series. S_n=a_1(1-r^n)/1-r In this formula, a_1 is the first term, r is the common ratio, and n is the number of terms. Let's start by paying close attention to the summation notation. ∑^9_(i= 1) 6 ( 7 )^(i-1) We can see that the common ratio of the series is 7. The lower limit is 1 and the upper limit is 9, so the number of terms is 9. Let's calculate the first term by substituting 1 for i into the indicated expression.
We are ready to evaluate the sum of the series. Let's substitute n= 9, r= 7, and a_1= 6 into the formula for the sum of a finite geometric series.
The sum of the series is 40 353 606.
Let's start by finding the common ratio of the geometric sequence. To do so, we can find the ratio between any two consecutive terms. For simplicity, we will find the ratio of the second term to the first term.
We found that the common ratio of our geometric sequence is 13. Recall that the absolute value of 13 is also 13. |1/3|=1/3 Since the absolute value of the common ratio is less than 1, the series converges.
From Part A, we already know that the series converges. To find its sum S, we will use the formula for the sum of an infinite geometric series.
S=a/1-r
Here, a is the first term of the series and r is the common ratio. We are given that the first term is 12 and we know from Part A that the common ratio is 13. Let's substitute these two values into the formula for the sum of an infinite geometric series and evaluate.
We found that the sum of the series is 34.
Let's start by finding the common ratio of the geometric sequence. To do so, we can find the ratio between any two consecutive terms. For simplicity, we will find the ratio of the second term to the first term.
We found that the common ratio of our geometric sequence is 1.01. Recall that the absolute value of 1.01 is also 1.01. |1.01|=1.01 Since the absolute value of the common ratio is greater than or equal to 1, the series diverges.
From Part A, we already know that the series diverges. Therefore, it is not possible to find its sum.
The series diverges.
Write the repeating decimal number as a fraction in its simplest form. 0.444...
We want to write the repeating decimal 0.444 ... as a fraction. To do so, we will start by writing this number as a sum of decimals and then as a sum of fractions.
| Number | 0.444 ... |
|---|---|
| Sum of Decimals | 0.4+0.04+0.004+... |
| Sum of Fractions | 4/10+4/100+4/1000+... |
Consider the sum of fractions above. Note that we can think of it as a geometric series with first term a_1= 410. To find its common ratio r, we can find the ratio of any two terms of the series. For simplicity, we will find the ratio of a_2 to a_1.
Each term of the series can be found by multiplying the previous term by the common ratio 110.
Let's substitute r= 110 and a_1= 410 in the formula for the sum of an infinite geometric series.
We found that the series formed by the sum of the terms is 49. With this information we can express the given number as a fraction. 0.444 ... =4/9
Consider the following infinite geometric series. 4+12/5+36/25+108/125+324/625+... Find the partial sum S_(12). Round the answer to two decimal places.
We are given an infinite geometric series and want to find the sum of its first 12 terms. To do so, let's start by recalling the formula for the partial sum of an infinite geometric series. S_n=a_1(1-r^n)/1-r Here, S_n represents the sum of the first n terms and a_1 and r are its first term and common ratio, respectively. Since we want to find the sum of the first 12 terms, we know that n= 12. We can also see that the first term is a_1= 4. To find the common ratio r, we can find the ratio between any two consecutive terms. For simplicity, we will find the ratio of the second term to the first term.
We found that the common ratio of the geometric sequence is r= 35. We can now use the values we have to find the sum of the first 12 terms. Let's do it!
The sum of the first 12 terms of the series is approximately 9.98.