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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.**

Consider the following sum of a sequence.

$21 +201 +2001 +20001 +⋯ $

a Is the sum finite or infinite?

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A geometric series is the sum of the terms of a geometric sequence.

$GeometricSequence1,2,4,8,… GeometricSeries1+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. $Sum ofInfinite Termsi=1∑∞ 2_{i−1} Sum ofnTermsi=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.

$S_{n}=1−ra_{1}(1−r_{n}) ,$ $r =1$

To derive the formula for a geometric series, a geometric sequence of $n$ terms with the first term $a_{1}$ and the common ratio $r$ will be considered.
Then, both sides of the equation will be multiplied by $(1−r)$ and the resulting equation simplified.
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.

$a_{1},a_{1}r,a_{1}r_{2},…,a_{1}r_{n−1} $

All the terms will be added to express the series and $a_{1}$ will be factored out.
$S_{n}=a_{1}+a_{1}r+a_{1}r_{2}+⋯+a_{1}r_{n−1}$

FactorOut

Factor out $a_{1}$

$S_{n}=a_{1}(1+r+r_{2}+⋯+r_{n−1})$

$S_{n}=a_{1}(1+r+r_{2}+⋯+r_{n−1})$

MultEqn

$LHS⋅(1−r)=RHS⋅(1−r)$

$S_{n}(1−r)=a_{1}(1+r+r_{2}+⋯+r_{n−1})(1−r)$

Distr

Distribute $(1−r)$

$S_{n}(1−r)=a_{1}(1(1−r)+r(1−r)+r_{2}(1−r)+⋯+r_{n−1}(1−r))$

Multiply

Multiply

$S_{n}(1−r)=a_{1}(1−r+r−r_{2}+r_{2}−r_{3}+⋯+r_{n−1}−r_{n})$

$S_{n}(1−r)=a_{1}(1−r+r−r_{2}+r_{2}−r_{3}+⋯+r_{n−2}−r_{n−1}+r_{n−1}−r_{n})$

AddSubTerms

Add and subtract terms

$S_{n}(1−r)=a_{1}(1−r_{n})$

DivEqn

$LHS/(1−r)=RHS/(1−r)$

$S_{n}=1−ra_{1}(1−r_{n}) $

$S_{n}=1−ra_{1}(1−r_{n}) $

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.
### Hint

### Solution

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 $n_{th}$ day.
The number of newly-infected students on the $n_{th}$ day can be represented by the formula $a_{n}=3_{n−1}.$
If the school does not suspend classes, there will be $1093$ infected students in $7$ days.

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.

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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a The number of newly-infected students each day makes a geometric sequence.

b Use the formula for the sum of a geometric series.

a It is given that the number of newly-infected students is triple the number of students that were infected the day before.

$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.
$a_{n}=a_{1}⋅r_{n−1}$

SubstituteII

$a_{1}=1$, $r=3$

$a_{n}=1⋅3_{n−1}$

IdPropMult

Identity Property of Multiplication

$a_{n}=3_{n−1}$

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.

$S_{n}=1−ra_{1}(1−r_{n}) $

Since the total number of infected students on the $7_{th}$ 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 $3.$
$S_{n}=1−ra_{1}(1−r_{n}) $

SubstituteValues

Substitute values

$S_{7}=1−31(1−3_{7}) $

Evaluate right-hand side

IdPropMult

Identity Property of Multiplication

$S_{7}=1−31−3_{7} $

ExpandFrac

$ba =b⋅(-1)a⋅(-1) $

$S_{7}=3−13_{7}−1 $

CalcPow

Calculate power

$S_{7}=3−12187−1 $

SubTerms

Subtract terms

$S_{7}=22186 $

CalcQuot

Calculate quotient

$S_{7}=1093$

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Sum:","formTextAfter":null,"answer":{"text":["4782960"]}}

Use the formula for the sum of a finite geometric series.

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.

$S_{n}=1−ra_{1}(1−r_{n}) ,r =1 $

In this formula, $a_{1}$ 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 $a_{1}.$
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_{1}a_{2} =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 $S_{n}$ and simplify.
$S_{n}=1−ra_{1}(1−r_{n}) $

SubstituteValues

Substitute values

$S_{12}=1−318(1−3_{12}) $

$S_{12}=4782960$

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.

$S_{∞}=1−ra_{1} ,$ $-1<r<1$

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.

To 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.
*very* small as the value of $n$ increases. In other words, it gets closer to $0$ as $n$ approaches infinity.
The formula for the sum of an infinite geometric series with $-1<r<1$ has been derived.

$S_{n}=1−ra_{1}(1−r_{n}) $

Since $r$ is a number between $-1$ and $1,$ the value of $r_{n}$ becomes $r_{n}1n→∞ 0 $

Therefore, $r_{n}=0$ can be substituted into the standard formula and the resulting equation simplified. $S_{n}=1−ra_{1}(1−r_{n}) $

SubstituteII

$r_{n}=0$, $n=∞$

$S_{∞}=1−ra_{1}(1−0) $

SubTerm

Subtract term

$S_{∞}=1−ra_{1}(1) $

IdPropMult

Identity Property of Multiplication

$S_{∞}=1−ra_{1} $

$S_{∞}=1−ra_{1} $

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.
### Answer

### Hint

### Solution

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.

a **Does the series converge or diverge?** Converge

**Sum**: $3$

b **Does the series converge or diverge?** Diverge

**Sum**: No sum.

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?

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−ra_{1} $

To find the sum, substitute $a_{1}=1$ and $r=32 $ into the formula and evaluate.
$S_{∞}=1−ra_{1} $

SubstituteII

$a_{1}=1$, $r=32 $

$S_{∞}=1−32 1 $

Evaluate right-hand side

OneToFrac

Rewrite $1$ as $33 $

$S_{∞}=33 −32 1 $

SubFrac

Subtract fractions

$S_{∞}=31 1 $

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.

$a_{1}a_{2} $

SubstituteII

$a_{1}=-410 $, $a_{2}=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.
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.

$S_{n}=1−ra_{1}(1−r_{n}) ,$ $r =1$

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.

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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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.

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: 3m3(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_{∞}=1−ra_{1} $

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 $n=3.$
$S_{n}=1−ra_{1}(1−r_{n}) $

With this in mind, substitute $n=3,$ $a_{1}=6(0.85),$ and $r=0.85$ into the formula and evaluate the result.
$S_{n}=1−ra_{1}(1−r_{n}) $

SubstituteValues

Substitute values

$S_{3}=1−0.856(0.85)(1−(0.85)_{3}) $

$S_{3}=13.11975$

$13.11975+3(0.85)_{4}$

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.
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?{"type":"choice","form":{"alts":["Tadeo","Tearrik"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

Recall that a series is geometric if it has a common ratio.

To determine who will save more money in $15$ months, the total amounts saved will be calculated one at a time.

$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}=1−ra_{1}(1−r_{n}) $

Now, substitute $n=15,$ $a_{1}=0.50,$ and $r=2$ into this formula and simplify.