Topics
4. Rational Functions and Equations
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United States of America
Grades 9-12

4.06 Geometric series

Lesson
Practice

Introduction

In Math 1 lesson  5.03 Geometric sequences  , we were introduced to the recursive pattern of multiplication in geometric sequences. In Math 1 lesson  5.04 Connecting sequences and functions  , we defined the sequences using an explicit rule and found terms within the sequence using that rule. In this lesson, we will derive a formula and use it to find the sum of a specific number of terms within a geometric sequence.

Finite geometric series

Recall a geometric sequence is a list of numbers in which each consecutive pair of numbers has a common ratio.

A table with 2 rows titled n and f of n, and with 4 columns. The data is as follows: First row: 0, 1, 2, and 3; Second row: 1, 3, 9, and 27. Below the second row are 3 semi circle arrows: from the cell containing 1 to the cell containing 3, from 3 to 9, and from 9 to 27. Below each semi circle arrows is a row of equations titled common ratio: below the left semi circle arrow, 3 divided 1 equals 3; below the middle semi circle arrow, 9 divided by 3 equals 3; below the right semi circle arrow, 27 divided by 3 equals 3.

The explicit rule for finding the nth term in a geometric sequence is

\displaystyle a_n=a_1\left(r\right)^{n-1}
\bm{a_n}
The nth term
\bm{a_1}
The first term
\bm{r}
The common ratio
\bm{n}
The number of the desired term

Exploration

Since his 1st birthday, Manuel has received cash gifts from various family members for each birthday. Each year, his mother takes \$50 of the money and puts it into a high-yield savings account earning 3\% interest annually.

On his first birthday, the account has \$50. On his second birthday, the money from his 1st birthday will accrue interest and another \$50 would get deposited. On his 2nd birthday, the money from his 1st birthday will accrue interest again, the money from his 2nd birthday will accrue interest, and another \$50 would be deposited. This pattern continues as shown in the table below.

BirthdayAmount in the account after each birthday
150
250\left(1.03\right)+50
350\left(1.03\right)^{2}+50\left(1.03\right)+50
450\left(1.03\right)^{3}+50\left(1.03\right)^{2}+50\left(1.03\right)+50
550\left(1.03\right)^{4}+50\left(1.03\right)^{3}+50\left(1.03\right)^{2}+50\left(1.03\right)+50

Suppose Manuel wants a way to easily determine the total amount of money in his account after any given birthday. Let S_n represent the amount of money in the account after the nth birthday. Use the questions below to derive a shorter formula for S_n.

  1. {S_5=50\left(1.03\right)^{4}+50\left(1.03\right)^{3}+50\left(1.03\right)^{2}+50\left(1.03\right)+50} represents the amount in the account after Manuel's 5th birthday. If we multiply both sides of the equation for S_5 by the common ratio, 1.03, how will the new equation compare to the equation for S_5?

  2. How could we use the given equation and the equation from question 1 to derive a shorter way for finding the sum of money in the account after 5 birthdays?

  3. Use the ideas from the previous questions to derive a formula for finding the total amount of money in Manuel's account after n birthdays.

A finite geometric series is the sum of the first n terms of a geometric sequence. It may also be called a partial sum of the sequence. We can find the sum of the first n terms by the formula

\displaystyle S_n=\frac{a_1\left(1-r^n\right)}{1-r}
\bm{a_1}
The first term of the geometric sequence
\bm{r}
The common ratio of the geometric sequence
\bm{n}
The number of terms being added

This can only be applied to geometric sequences where the common ratio is not 1 because r=1 would make the equation undefined.

Recall from lesson  3.02 The binomial theorem  that we can use a capital sigma to represent a summation. A geometric series can be represented in summation notation by \sum_{1}^n a\left(r\right)^{n-1}=a+ar+ar^2+\cdots+ar^{n-1}This notation allows us to identify key information about the individual terms in the sequence, which can then be used in the formula to quickly calculate the final sum if n is large.

Examples

Example 1

Sissa is a mythical character from India. Sissa was invited to meet with the king after inventing the game of Chaturanga, an ancient Indian game that evolved into the modern games of chess, xiangqi, shogi, and others.

The king, delighted with his creation, asked Sissa how he wanted to be rewarded. Sissa started placing grain in a pattern on the board as shown in the image below.

A figure of two persons on opposite sides of an 8 by 8 checkerboard. On the side facing one of the players, the number of grains are placed starting from the leftmost square of the closest row to the player: 1, 2, 4, and 8. A right arrow is placed on the fifth square.
a

Assuming this pattern continues, derive a formula to find the total number of grains of rice on the first n tiles of the board.

Worked Solution
Create a strategy

The geometric sequence is 1,\,2,\,4,\,8,\,16,\,32,\,\ldots with a common ratio of 2. Recall that any term in the finite sequence will be in the form a_n=a_1\left(2\right)^{n-1}. Since the first term of this sequence is 1, the last term will be 2^{n-1}. A geometric series is the sum of all the terms, so we have S_{n}=1+2+4+8+16+\cdots+2^{n-1}.

Apply the idea

If we multiply this series by 2, the common ratio, and subtract the new series from the original one, most of the terms will eliminate.

\begin{aligned}S_{n}&=1+2+4+8+16+\cdots+2^{n-1}\\2S_{n}&=2+4+8+16+32+\cdots+2^{n}\end{aligned}

\begin{aligned}S_{n}&=1+2+4+8+16+\cdots+2^{n-1}\\-2S_{n}&=\text{ }\text{ }-2-4-8-16-32-\cdots-2^{n}\\\hline S_{n}-2S_{n}&=1-2^{n}\end{aligned}

Next, we can factor out S_n from the left side of the equation and solve for it by dividing by what is left in parentheses.

\begin{aligned}S_{n}\left(1-2\right)&=1-2^{n}\\S_n\left(-1\right)&=1-2^n\\S_n&=2^n-1\end{aligned}

b

Use the formula you derived in part (a) to find the total number of grains of rice on the first 2 rows of the board.

Worked Solution
Create a strategy

There are 16 tiles in the first 2 rows, so we will use n=16 with the formula we derived.

Apply the idea
\displaystyle S_n\displaystyle =\displaystyle 2^n-1Formula derived in part (a)
\displaystyle S_{16}\displaystyle =\displaystyle 2^{16}-1Substitute n=16
\displaystyle =\displaystyle 65\,535Evaluate
Reflect and check

If you were the king in this situation, would you have agreed to Sissa's reward proposal? After how many tiles would at least 1 million grains of rice have accumulated?

Example 2

Derive the formula for any geometric series given the first term, a, and the common ratio, r.

Worked Solution
Create a strategy

We can use a strategy similar to what we did in previous question 1, but replace the common ratio with r and the number of terms being added with n. The series would be S_n=a+ar+ar^2+ar^3+\cdots +ar^{n-1}.

Apply the idea

If we multiply both sides by the common ratio, r, we get rS_n=ar+ar^2+ar^3+ar^4+\cdots +ar^{n}Subtracting this from the original expression, we see that the middle terms eliminate each other \begin{aligned}S_{n}&=a+ar+ar^2+ar^3+ar^4+\cdots+ar^{n-1}\\-rS_{n}&=\text{ }\text{ }-ar-ar^2-ar^3-ar^4-ar^5-\cdots-ar^{n}\\\hline S_{n}-rS_{n}&=a-ar^{n}\end{aligned} Factoring S_n from the lefthand side and a from the righthand side, we get S_n\left(1-r\right)=a\left(1-r^{n}\right) Finally, we can restrict the domain so that r\neq 1 and divide both sides by 1-r to get S_n=\dfrac{a\left(1-r^n\right)}{1-r} where a is the initial term, r is the common ratio, and n is the number of terms being added.

Reflect and check

Another way to derive the formula is by using an identity we learned in  3.03 Polynomial identities  . If we factor a from the right side of the equation, we get S_n=a\left(1+r+r^2+r^3+\cdots +r^{n-1}\right) Next, we will multiply both sides by r-1 in order to use the identity x^{n}-1=\left(x-1\right)\left(x^{n-1}+x^{n-2}+\ldots+x+1\right).

\displaystyle \left(r-1\right)S_{n}\displaystyle =\displaystyle a\left(r-1\right)\left(r^{n-1}+\cdots+r^2+r+1\right)Multiply both sides by r-1
\displaystyle \left(r-1\right)S_{n}\displaystyle =\displaystyle a\left(r^{n}-1\right)Use the stated identity
\displaystyle S_{n}\displaystyle =\displaystyle \frac{a\left(r^{n}-1\right)}{r-1}Divide both sides by r-1
\displaystyle S_n\displaystyle =\displaystyle \frac{-a\left(1-r^{n}\right)}{-1\left(1-r\right)}Factor out -1
\displaystyle S_{n}\displaystyle =\displaystyle \frac{a\left(1-r^n\right)}{1-r}Simplify since \frac{-1}{-1}=1

Example 3

Find the sum of the first 9 terms of the geometric sequence 2, -\dfrac{3}{2},\dfrac{9}{8}, -\dfrac{27}{32},\ldots

Worked Solution
Create a strategy

To use the geometric series formula, we need to know the first term, the common ratio, and the number of terms being added. The first term and the number of terms being added is given to us. The common ratio can be found by dividing any term by the previous term.

Apply the idea

The common ratio is

\displaystyle r\displaystyle =\displaystyle \dfrac{9}{8}\div-\dfrac{3}{2}
\displaystyle =\displaystyle \dfrac{9}{8}\times-\dfrac{2}{3}
\displaystyle =\displaystyle -\dfrac{3}{4}

Now, we can find the sum of the first 9 terms using the geometric series formula.

\displaystyle S_n\displaystyle =\displaystyle \dfrac{a_1\left(1-r^n\right)}{1-r}Geometric series formula
\displaystyle S_{9}\displaystyle =\displaystyle \dfrac{2\left(1-\left(-\frac{3}{4}\right)^{9}\right)}{1-\left(-\frac{3}{4}\right)}Substitute n=9,a_1=2, r=-\frac{3}{4}
\displaystyle =\displaystyle \dfrac{2\left(1-\left(-\frac{19683}{262144}\right)\right)}{1-\left(-\frac{3}{4}\right)}Evaluate the exponent
\displaystyle =\displaystyle \dfrac{2\left(\frac{281827}{262144}\right)}{\frac{7}{4}}Evaluate the addition
\displaystyle =\displaystyle \dfrac{\frac{281827}{131072}}{\frac{7}{4}}Evaluate the multiplication
\displaystyle =\displaystyle \dfrac{281827}{262144}\times\dfrac{4}{7}Evaluate the division
\displaystyle =\displaystyle \dfrac{40261}{32768}Evaluate the multiplication

Using a calculator, we can approximate the sum to be about 1.229.

Reflect and check

This geometric sequence has the rule a_n=2\left(-\dfrac{3}{4}\right)^{n-1} If we want to add the first 9 terms, the sigma notation would be \sum_{n=1}^{9} 2\left(-\dfrac{3}{4}\right)^{n-1}

This notation allows us to see each individual term, but we can identify the key information from this notation and use the geometric series formula instead.

Example 4

Erin drinks 35\text{ mg} of caffeine every 8 hours. She has coffee in the morning, cola mid-afternoon, and tea before bed. If 20\% of the caffeine is still in Erin's body after 8 hours, determine how much caffeine is her body after 3 days.

Worked Solution
Create a strategy

To see how the caffeine accumulates during the first day, we can write out the first 3 terms of the series.

In the morning, Erin has 35\text{ mg} of caffeine in her coffee. After 8 hours, 20\% of that caffeine is left in her body and she adds 35\text{ mg} more by drinking a cola. 35\left(0.2\right)+35 In the evening, 20\% of the 20\% of caffeine from her morning coffee remains, 20\% of her mid-afternoon cola remains, and she drinks another 35\text{ mg} in her tea. 35\left(0.2\right)^2+35\left(0.2\right)+35

This pattern will continue, and we can see that the terms form a geometric sequence defined by {a_n= 35\left(0.2\right)^{n-1}} where n represents the number of 8-hour periods that have passed.

Apply the idea

To find out how much caffeine is in Erin's body after 3 days, we will need to use n=9 since there are nine 8-hour periods in 3 days. Using a_1=35,r=0.2,n=9 with the geometric series formula, we get:

\displaystyle S_9\displaystyle =\displaystyle \dfrac{35\left(1-0.2^9\right)}{1-0.2}
\displaystyle \approx\displaystyle 43.75\text{ mg}

About 43.75\text{ mg} of caffeine is left in Erin's body after 3 days of drinking 35\text{ mg} every 8 hours.

Reflect and check

In reality, there is likely less than this amount of caffeine in Erin's body after 3 days. Caffeine can be completely flushed from your body after about 10 hours if you drink enough water. In this model, the caffeine from the coffee on the first day will never entirely leave Erin's body.

Idea summary

We can find the sum of the first n terms of a geometric sequence by the geometric series formula

\displaystyle S_n=\frac{a_1\left(1-r^n\right)}{1-r}
\bm{a_1}
The first term of the geometric sequence
\bm{r}
The common ratio of the geometric sequence
\bm{n}
The number of terms being added

Outcomes

A.SSE.B.4

Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems.

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