Series and Their Notations

Learning Objectives

In this section, you will:

• Use summation notation.
• Use the formula for the sum of the first n terms of an arithmetic series.
• Use the formula for the sum of the first n terms of a geometric series.
• Use the formula for the sum of an infinite geometric series.
• Solve annuity problems.

A parent decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned.

Using Summation Notation

To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a series. Consider, for example, the following series.

3+7+11+15+19+…3+7+11+15+19+…

The nth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation Sn𝑆𝑛 represents the partial sum.

S1=3S2=3+7=10S3=3+7+11=21S4=3+7+11+15=36𝑆1=3𝑆2=3+7=10𝑆3=3+7+11=21𝑆4=3+7+11+15=36

Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, Σ,Σ, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation, is the number used to generate the last term in a series.

If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ak=2k𝑎𝑘=2𝑘 for k=1𝑘=1 through k=5.𝑘=5. We can begin by substituting the terms for k𝑘 and listing out the terms of this series.

a1=2(1)=2a2=2(2)=4a3=2(3)=6a4=2(4)=8a5=2(5)=10𝑎1=2(1)=2𝑎2=2(2)=4𝑎3=2(3)=6𝑎4=2(4)=8𝑎5=2(5)=10

We can find the sum of the series by adding the terms:

∑k=152k=2+4+6+8+10=30∑𝑘=152𝑘=2+4+6+8+10=30

SUMMATION NOTATION

The sum of the first n𝑛 terms of a series can be expressed in summation notation as follows:

∑k=1nak∑𝑘=1𝑛𝑎𝑘

This notation tells us to find the sum of ak𝑎𝑘 from k=1𝑘=1 to k=n.𝑘=𝑛.

k𝑘 is called the index of summation, 1 is the lower limit of summation, and n𝑛 is the upper limit of summation.

Q&A

Does the lower limit of summation have to be 1?

No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.

HOW TO

Given summation notation for a series, evaluate the value.

1. Identify the lower limit of summation.
2. Identify the upper limit of summation.
3. Substitute each value of k𝑘 from the lower limit to the upper limit into the formula.
4. Add to find the sum.

EXAMPLE 1

Using Summation Notation

Evaluate ∑k=37k2.∑𝑘=37𝑘2.

Solution

According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of k2𝑘2 from k=3𝑘=3 to k=7.𝑘=7. We find the terms of the series by substituting k=3,4,5,6,𝑘=3,4,5,6, and 77 into the function k2.𝑘2. We add the terms to find the sum.

∑k=37k2=32+42+52+62+72=9+16+25+36+49=135∑𝑘=37𝑘2=32+42+52+62+72=9+16+25+36+49=135

TRY IT #1

Evaluate ∑k=25(3k–1).∑𝑘=25(3𝑘–1).

Using the Formula for Arithmetic Series

Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, d.𝑑. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first n𝑛 terms of an arithmetic series as:

Sn=a1+(a1+d)+(a1+2d)+…+(an–d)+an.𝑆𝑛=𝑎1+(𝑎1+𝑑)+(𝑎1+2𝑑)+…+(𝑎𝑛–𝑑)+𝑎𝑛.

We can also reverse the order of the terms and write the sum as

Sn=an+(an–d)+(an–2d)+…+(a1+d)+a1.𝑆𝑛=𝑎𝑛+(𝑎𝑛–𝑑)+(𝑎𝑛–2𝑑)+…+(𝑎1+𝑑)+𝑎1.

If we add these two expressions for the sum of the first n𝑛 terms of an arithmetic series, we can derive a formula for the sum of the first n𝑛 terms of any arithmetic series.

Sn=a1+(a1+d)+(a1+2d)+…+(an–d)+an+Sn=an+(an–d)+(an–2d)+…+(a1+d)+a12Sn=(a1+an)+(a1+an)+…+(a1+an)𝑆𝑛=𝑎1+(𝑎1+𝑑)+(𝑎1+2𝑑)+…+(𝑎𝑛–𝑑)+𝑎𝑛+𝑆𝑛=𝑎𝑛+(𝑎𝑛–𝑑)+(𝑎𝑛–2𝑑)+…+(𝑎1+𝑑)+𝑎12𝑆𝑛=(𝑎1+𝑎𝑛)+(𝑎1+𝑎𝑛)+…+(𝑎1+𝑎𝑛)

Because there are n𝑛 terms in the series, we can simplify this sum to

2Sn=n(a1+an).2𝑆𝑛=𝑛(𝑎1+𝑎𝑛).

We divide by 2 to find the formula for the sum of the first n𝑛 terms of an arithmetic series.

Sn=n(a1+an)2𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2

FORMULA FOR THE SUM OF THE FIRST N TERMS OF AN ARITHMETIC SERIES

An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first n𝑛 terms of an arithmetic sequence is

Sn=n(a1+an)2𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2

HOW TO

Given terms of an arithmetic series, find the sum of the first n𝑛 terms.

1. Identify a1𝑎1 and an.𝑎𝑛.
2. Determine n.𝑛.
3. Substitute values for a1, an,𝑎1, 𝑎𝑛, and n𝑛 into the formula Sn=n(a1+an)2.𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2.
4. Simplify to find Sn.𝑆𝑛.

EXAMPLE 2

Finding the First n Terms of an Arithmetic Series

Find the sum of each arithmetic series.

1. ⓐ 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 325 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32
2. ⓑ 20 + 15 + 10 +…+ −5020 + 15 + 10 +…+ −50
3. ⓒ ∑k=1123k−8∑𝑘=1123𝑘−8

Solution

1. ⓐWe are given a1=5𝑎1=5 and an=32.𝑎𝑛=32.Count the number of terms in the sequence to find n=10.𝑛=10.Substitute values for a1,an,𝑎1,𝑎𝑛, and n𝑛 into the formula and simplify. Sn=n(a1+an)2S10=10(5+32)2=185 𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2𝑆10=10(5+32)2=185
2. ⓑWe are given a1=20𝑎1=20 and an=−50.𝑎𝑛=−50.Use the formula for the general term of an arithmetic sequence to find n.𝑛.an=a1+(n−1)d−50=20+(n−1)(−5)−70=(n−1)(−5)14=n−115=n𝑎𝑛=𝑎1+(𝑛−1)𝑑−50=20+(𝑛−1)(−5)−70=(𝑛−1)(−5)14=𝑛−115=𝑛Substitute values for a1,an,n𝑎1,𝑎𝑛,𝑛 into the formula and simplify.
   Sn=n(a1+an)2S15=15(20−50)2=−225𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2𝑆15=15(20−50)2=−225
3. ⓒTo find a1,𝑎1, substitute k=1𝑘=1 into the given explicit formula.ak=3k−8a1=3(1)−8=−5𝑎𝑘=3𝑘−8𝑎1=3(1)−8=−5We are given that n=12.𝑛=12. To find a12,𝑎12, substitute k=12𝑘=12 into the given explicit formula.ak=3k−8a12=3(12)−8=28𝑎𝑘=3𝑘−8𝑎12=3(12)−8=28Substitute values for a1,an,𝑎1,𝑎𝑛, and n𝑛 into the formula and simplify.Sn=n(a1+an)2S12=12(−5+28)2=138𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2𝑆12=12(−5+28)2=138

Use the formula to find the sum of each arithmetic series.

TRY IT #2

1.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.41.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.4

TRY IT #3

13 + 21 + 29 + …+ 6913 + 21 + 29 + …+ 69

TRY IT #4

∑k=1105−6k∑𝑘=1105−6𝑘

EXAMPLE 3

Solving Application Problems with Arithmetic Series

On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?

Solution

This problem can be modeled by an arithmetic series with a1=12𝑎1=12 and d=14.𝑑=14. We are looking for the total number of miles walked after 8 weeks, so we know that n=8,𝑛=8, and we are looking for S8.𝑆8. To find a8,𝑎8, we can use the explicit formula for an arithmetic sequence.

an=a1+d(n−1)a8=12+14(8−1)=94𝑎𝑛=𝑎1+𝑑(𝑛−1)𝑎8=12+14(8−1)=94

We can now use the formula for arithmetic series.

 Sn=n(a1+an)2 S8=8(12+94)2=11 𝑆𝑛=𝑛(𝑎1+𝑎𝑛)2 𝑆8=8(12+94)2=11

She will have walked a total of 11 miles.

TRY IT #5

A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?

Using the Formula for Geometric Series

Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, r.𝑟. We can write the sum of the first n𝑛 terms of a geometric series as

Sn=a1+ra1+r2a1+…+rn–1a1.𝑆𝑛=𝑎1+𝑟𝑎1+𝑟2𝑎1+…+𝑟𝑛–1𝑎1.

Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first n𝑛 terms of a geometric series. We will begin by multiplying both sides of the equation by r.𝑟.

rSn=ra1+r2a1+r3a1+…+rna1𝑟𝑆𝑛=𝑟𝑎1+𝑟2𝑎1+𝑟3𝑎1+…+𝑟𝑛𝑎1

Next, we subtract this equation from the original equation.

   Sn=a1+ra1+r2a1+…+rn–1a1−rSn=−(ra1+r2a1+r3a1+…+rna1)(1−r)Sn=a1−rna1   𝑆𝑛=𝑎1+𝑟𝑎1+𝑟2𝑎1+…+𝑟𝑛–1𝑎1−𝑟𝑆𝑛=−(𝑟𝑎1+𝑟2𝑎1+𝑟3𝑎1+…+𝑟𝑛𝑎1)(1−𝑟)𝑆𝑛=𝑎1−𝑟𝑛𝑎1

Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for Sn,𝑆𝑛, divide both sides by (1−r).(1−𝑟).

Sn=a1(1−rn)1−rr≠1𝑆𝑛=𝑎1(1−𝑟𝑛)1−𝑟r≠1

FORMULA FOR THE SUM OF THE FIRST N TERMS OF A GEOMETRIC SERIES

A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first n𝑛 terms of a geometric sequence is represented as

Sn=a1(1−rn)1−rr≠1𝑆𝑛=𝑎1(1−𝑟𝑛)1−𝑟r≠1

HOW TO

Given a geometric series, find the sum of the first n terms.

1. Identify a1,r,andn.𝑎1,𝑟,and𝑛.
2. Substitute values for a1,r,𝑎1,𝑟, and n𝑛 into the formula Sn=a1(1–rn)1–r.𝑆𝑛=𝑎1(1–𝑟𝑛)1–𝑟.
3. Simplify to find Sn.𝑆𝑛.

EXAMPLE 4

Finding the First n Terms of a Geometric Series

Use the formula to find the indicated partial sum of each geometric series.

1. ⓐ S11𝑆11 for the series 8 + -4 + 2 + …8 + -4 + 2 + …
2. ⓑ ∑​6k=13⋅2k∑​𝑘=163⋅2𝑘

Solution

1. ⓐa1=8,𝑎1=8, and we are given that n=11.𝑛=11.We can find r𝑟 by dividing the second term of the series by the first.r=−48=−12𝑟=−48=−12Substitute values for a1, r, and n𝑎1, 𝑟, and 𝑛 into the formula and simplify.Sn=a1(1−rn)1−rS11=8(1−(−12)11)1−(−12)≈5.336𝑆𝑛=𝑎1(1−𝑟𝑛)1−𝑟𝑆11=8(1−(−12)11)1−(−12)≈5.336
2. ⓑFind a1𝑎1 by substituting k=1𝑘=1 into the given explicit formula.a1=3⋅21=6𝑎1=3⋅21=6We can see from the given explicit formula that r=2.𝑟=2. The upper limit of summation is 6, so n=6.𝑛=6.Substitute values for a1,r,𝑎1,𝑟, and n𝑛 into the formula, and simplify.Sn=a1(1−rn)1−rS6=6(1−26)1−2=378𝑆𝑛=𝑎1(1−𝑟𝑛)1−𝑟𝑆6=6(1−26)1−2=378

Use the formula to find the indicated partial sum of each geometric series.

TRY IT #6

S20𝑆20 for the series 1,000 + 500 + 250 + …1,000 + 500 + 250 + …

TRY IT #7

∑k=183k∑𝑘=183𝑘

EXAMPLE 5

Solving an Application Problem with a Geometric Series

At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.

Solution

The problem can be represented by a geometric series with a1=26,750;𝑎1=26,750; n=5;𝑛=5; and r=1.016.𝑟=1.016. Substitute values for a1,𝑎1, r,𝑟, and n𝑛 into the formula and simplify to find the total amount earned at the end of 5 years.

Sn=a1(1−rn)1−rS5=26,750(1−1.0165)1−1.016≈138,099.03𝑆𝑛=𝑎1(1−𝑟𝑛)1−𝑟𝑆5=26,750(1−1.0165)1−1.016≈138,099.03

He will have earned a total of $138,099.03 by the end of 5 years.

TRY IT #8

At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?

Using the Formula for the Sum of an Infinite Geometric Series

Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first n𝑛 terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is 2+4+6+8+…2+4+6+8+…

This series can also be written in summation notation as ∑k=1∞2k,∑𝑘=1∞2𝑘, where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges.

Determining Whether the Sum of an Infinite Geometric Series is Defined

If the terms of an infinite geometric sequence approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:

1+0.2+0.04+0.008+0.0016+…1+0.2+0.04+0.008+0.0016+…

The common ratio r= 0.2.𝑟= 0.2. As n𝑛 gets very large, the values of rn𝑟𝑛 get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with −1<r<1−1<𝑟<1 approach 0; the sum of a geometric series is defined when −1<r<1.−1<𝑟<1.

DETERMINING WHETHER THE SUM OF AN INFINITE GEOMETRIC SERIES IS DEFINED

The sum of an infinite series is defined if the series is geometric and −1<r<1.−1<𝑟<1.

HOW TO

Given the first several terms of an infinite series, determine if the sum of the series exists.

1. Find the ratio of the second term to the first term.
2. Find the ratio of the third term to the second term.
3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.
4. If a common ratio, r,𝑟, was found in step 3, check to see if −1<r<1−1<𝑟<1 . If so, the sum is defined. If not, the sum is not defined.

EXAMPLE 6

Determining Whether the Sum of an Infinite Series is Defined

Determine whether the sum of each infinite series is defined.

1. ⓐ 12 + 8 + 4 + …12 + 8 + 4 + …
2. ⓑ 34+12+13+…34+12+13+…
3. ⓒ ∑∞k=127⋅(13)k∑𝑘=1∞27⋅(13)𝑘
4. ⓓ ∑k=1∞5k∑𝑘=1∞5𝑘

Solution

1. ⓐThe ratio of the second term to the first is 23,23, which is not the same as the ratio of the third term to the second, 12.12. The series is not geometric.
2. ⓑThe ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of 23.23. The sum of the infinite series is defined.
3. ⓒThe given formula is exponential with a base of 13;13; the series is geometric with a common ratio of 13.13. The sum of the infinite series is defined.
4. ⓓThe given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.

Determine whether the sum of the infinite series is defined.

TRY IT #9

13+12+34+98+…13+12+34+98+…

TRY IT #10

24+(−12)+6+(−3)+…24+(−12)+6+(−3)+…

TRY IT #11

∑∞k=115⋅(–0.3)k∑𝑘=1∞15⋅(–0.3)𝑘

Finding Sums of Infinite Series

When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first n𝑛 terms of a geometric series.

Sn=a1(1−rn)1−r𝑆𝑛=𝑎1(1−𝑟𝑛)1−𝑟

We will examine an infinite series with r=12.𝑟=12. What happens to rn𝑟𝑛 as n𝑛 increases?

(12)2=14(12)3=18(12)4=116(12)2=14(12)3=18(12)4=116

The value of rn𝑟𝑛 decreases rapidly. What happens for greater values of n?𝑛?

(12)10=11,024(12)20=11,048,576(12)30=11,073,741,824(12)10=11,024(12)20=11,048,576(12)30=11,073,741,824

As n𝑛 gets very large, rn𝑟𝑛 gets very small. We say that, as n𝑛 increases without bound, rn𝑟𝑛 approaches 0. As rn𝑟𝑛 approaches 0, 1−rn1−𝑟𝑛 approaches 1. When this happens, the numerator approaches a1.𝑎1. This give us a formula for the sum of an infinite geometric series.

FORMULA FOR THE SUM OF AN INFINITE GEOMETRIC SERIES

The formula for the sum of an infinite geometric series with −1<r<1−1<𝑟<1 is

S=a11−r𝑆=𝑎11−𝑟

HOW TO

Given an infinite geometric series, find its sum.

1. Identify a1𝑎1 and r.𝑟.
2. Confirm that –1<r<1.–1<𝑟<1.
3. Substitute values for a1𝑎1 and r𝑟 into the formula, S=a11−r.𝑆=𝑎11−𝑟.
4. Simplify to find S.𝑆.

EXAMPLE 7

Finding the Sum of an Infinite Geometric Series

Find the sum, if it exists, for the following:

1. ⓐ 10+9+8+7+…10+9+8+7+…
2. ⓑ 248.6+99.44+39.776+…248.6+99.44+39.776+…
3. ⓒ ∑∞k=14,374⋅(–13)k–1∑𝑘=1∞4,374⋅(–13)𝑘–1
4. ⓓ ∑∞k=119⋅(43)k∑𝑘=1∞19⋅(43)𝑘

Solution

1. ⓐThere is not a constant ratio; the series is not geometric.
2. ⓑThere is a constant ratio; the series is geometric. a1=248.6𝑎1=248.6 and r=99.44248.6=0.4,𝑟=99.44248.6=0.4, so the sum exists. Substitute a1=248.6𝑎1=248.6 and r=0.4𝑟=0.4 into the formula and simplify to find the sum:S=a11−rS=248.61−0.4=414.3¯𝑆=𝑎11−𝑟𝑆=248.61−0.4=414.3¯
3. ⓒThe formula is exponential, so the series is geometric with r=–13.𝑟=–13. Find a1𝑎1 by substituting k=1𝑘=1 into the given explicit formula:a1=4,374⋅(–13)1–1=4,374𝑎1=4,374⋅(–13)1–1=4,374Substitute a1=4,374𝑎1=4,374 and r=−13𝑟=−13 into the formula, and simplify to find the sum:S=a11−rS=4,3741−(−13)=3,280.5𝑆=𝑎11−𝑟𝑆=4,3741−(−13)=3,280.5
4. ⓓThe formula is exponential, so the series is geometric, but r>1.𝑟>1. The sum does not exist.

EXAMPLE 8

Finding an Equivalent Fraction for a Repeating Decimal

Find an equivalent fraction for the repeating decimal 0.3¯0.3¯

Solution

We notice the repeating decimal 0.3¯=0.333…0.3¯=0.333… so we can rewrite the repeating decimal as a sum of terms.

0.3¯=0.3+0.03+0.003+…0.3¯=0.3+0.03+0.003+…

Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term.

Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have

Sn=a11−r=0.31−0.1=0.30.9=13.𝑆𝑛=𝑎11−𝑟=0.31−0.1=0.30.9=13.

Find the sum, if it exists.

TRY IT #12

2+23+29+…2+23+29+…

TRY IT #13

∑k=1∞0.76k+1∑𝑘=1∞0.76𝑘+1

TRY IT #14

∑k=1∞(−38)k∑𝑘=1∞(−38)𝑘

Solving Annuity Problems

At the beginning of the section, we looked at a problem in which a parent invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the parent invests $50 each month. This is the value of the initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added.

We can find the value of the annuity right after the last deposit by using a geometric series with a1=50𝑎1=50 and r=100.5%=1.005.𝑟=100.5%=1.005. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned.

We can find the value of the annuity after n𝑛 deposits using the formula for the sum of the first n𝑛 terms of a geometric series. In 6 years, there are 72 months, so n=72.𝑛=72. We can substitute a1=50, r=1.005, and n=72𝑎1=50, 𝑟=1.005, and 𝑛=72 into the formula, and simplify to find the value of the annuity after 6 years.

S72=50(1−1.00572)1−1.005≈4,320.44𝑆72=50(1−1.00572)1−1.005≈4,320.44

After the last deposit, the parent will have a total of $4,320.44 in the account. Notice, the parent made 72 payments of $50 each for a total of 72(50) = $3,600.72(50) = $3,600. This means that because of the annuity, the parent earned $720.44 interest in their college fund.

HOW TO

Given an initial deposit and an interest rate, find the value of an annuity.

1. Determine a1,𝑎1, the value of the initial deposit.
2. Determine n,𝑛, the number of deposits.
3. Determine r.𝑟.
   1. Divide the annual interest rate by the number of times per year that interest is compounded.
   2. Add 1 to this amount to find r.𝑟.
4. Substitute values for a1,r,andn𝑎1,𝑟,and𝑛 into the formula for the sum of the first n𝑛 terms of a geometric series, Sn=a1(1–rn)1–r.𝑆𝑛=𝑎1(1–𝑟𝑛)1–𝑟.
5. Simplify to find Sn,𝑆𝑛, the value of the annuity after n𝑛 deposits.

EXAMPLE 9

Solving an Annuity Problem

A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?

Solution

The value of the initial deposit is $100, so a1=100.𝑎1=100. A total of 120 monthly deposits are made in the 10 years, so n=120.𝑛=120. To find r,𝑟, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.

r=1+0.0912=1.0075𝑟=1+0.0912=1.0075

Substitute a1=100,r=1.0075,andn=120𝑎1=100,𝑟=1.0075,and𝑛=120 into the formula for the sum of the first n𝑛 terms of a geometric series, and simplify to find the value of the annuity.

S120=100(1−1.0075120)1−1.0075≈19,351.43𝑆120=100(1−1.0075120)1−1.0075≈19,351.43

So the account has $19,351.43 after the last deposit is made.

TRY IT #15

At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?