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Series and Their Notations

Use summation notation.

Series and Their Notations

Use summation notation to write a sum. (IA 12.1.5)
Find the sum of the first n terms of an arithmetic sequence. (IA 12.2.3)

A series is the sum of the terms of a sequence. For example, 1 + 6 + 11+ 16 + 21 + 26 + 31 is the sum of the first seven terms arithmetic sequence with general term, ${a}_{n}=5n-4.$

We write a series by using the summation notation. In order to write that summation, we will need to find the general term of our sequence and the summation will look like:

For the series, 1 + 6 + 11 + 16 + 21 + 26 + 31 + .... the summation notation is $\sum _{n=1}^{7}5n-4$

Example

Use summation notation to write the sum.

Try it.

Write the sum using summation notation: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$

Solution

	$\begin{array}{llllll}n: & 1, & 2, & 3, & 4, & 5\end{array}$
Look for a pattern in the terms.	Terms: $\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}$
The numerators are all one. The denominators are the counting numbers from 1 thru 5.	Pattern: $\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}$
The general terms is:	$\frac{1}{n}$
The sum in summation notation is:	$\sum _{n=1}^{5}\frac{1}{n}$

Use summation notation to write the sum.

Try it.

Write the sum using summation notation: $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$

Try it.

Write the sum using summation notation: $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}$

Try it.

Find the sum of the first 30 terms of the arithmetic sequence: 16, 10, 4, –2, –8, ………

Try it.

Find the sum of the first 50 terms of the arithmetic sequence whose general term is ${a}_{n}=2n+7$ .

Try it.

Find the sum: $\sum _{i=1}^{30}(7i-5)$

Condensed — the full section is in OpenStax College Algebra 2e.

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.

\[3+7+11+15+19+...\]

The nth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation ${S}_{n}$ represents the partial sum.

\[\begin{array}{l}{S}_{1}=3 \\ {S}_{2}=3+7=10 \\ {S}_{3}=3+7+11=21 \\ {S}_{4}=3+7+11+15=36\end{array}\]

Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, $\text{\Sigma },$ 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 ${a}_{k}=2k$ for $k=1$ through $k=5.$ We can begin by substituting the terms for $k$ and listing out the terms of this series.

\[\begin{array}{l}\begin{array}{l} \\ {a}_{1}=2(1)=2\end{array} \\ {a}_{2}=2(2)=4 \\ {a}_{3}=2(3)=6 \\ {a}_{4}=2(4)=8 \\ {a}_{5}=2(5)=10\end{array}\]

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

\[\sum _{k=1}^{5}2k=2+4+6+8+10=30\]

Condensed — the full section is in OpenStax College Algebra 2e.

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, $d.$ The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first $n$ terms of an arithmetic series as:

\[{S}_{n}={a}_{1}+({a}_{1}+d)+({a}_{1}+2d)+...+({a}_{n}-d)+{a}_{n}.\]

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

\[{S}_{n}={a}_{n}+({a}_{n}-d)+({a}_{n}-2d)+...+({a}_{1}+d)+{a}_{1}.\]

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

\[\frac{\begin{array}{l}\ \ \ {S}_{n}={a}_{1}+({a}_{1}+d)+({a}_{1}+2d)+...+({a}_{n}-d)+{a}_{n} \\ +\ {S}_{n}={a}_{n}+({a}_{n}-d)+({a}_{n}-2d)+...+({a}_{1}+d)+{a}_{1}\end{array}}{2{S}_{n}=({a}_{1}+{a}_{n})+({a}_{1}+{a}_{n})+...+({a}_{1}+{a}_{n})}\]

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

\[2{S}_{n}=n({a}_{1}+{a}_{n}).\]

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

\[{S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\]

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

Condensed — the full section is in OpenStax College Algebra 2e.

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, $r.$ We can write the sum of the first $n$ terms of a geometric series as

\[{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n-1}{a}_{1}.\]

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

\[r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\]

Next, we subtract this equation from the original equation.

\[\begin{array}{l} \\ \frac{\begin{array}{l}\ {S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n-1}{a}_{1} \\ -r{S}_{n}=-(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1})\end{array}}{(1-r){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\end{array}\]

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 ${S}_{n},$ divide both sides by $(1-r).$

\[{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r}\ \text{r}\ne \text{1}\]

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

Example

Try it.

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 ${a}_{1}=26,750\text{;}$ $n=5\text{;}$ and $r=1.016.$ Substitute values for ${a}_{1}\text{,}$ $r\text{,}$ and $n$ into the formula and simplify to find the total amount earned at the end of 5 years.

\[\begin{array}{l}{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r} \\ {S}_{5}=\frac{26\text{,}750(1-{1.016}^{5})}{1-1.016}\approx 138\text{,}099.03\end{array}\]

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

Condensed — the full section is in OpenStax College Algebra 2e.

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 $n$ terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is $2+4+6+8+...$

This series can also be written in summation notation as $\sum _{k=1}^{\infty }2k,$ 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.

Condensed — the full section is in OpenStax College Algebra 2e.

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 ${a}_{1}=50$ and $r=100.5\%=1.005.$ 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 $n$ deposits using the formula for the sum of the first $n$ terms of a geometric series. In 6 years, there are 72 months, so $n=72.$ We can substitute ${a}_{1}=50,r=1.005,\text{and}n=72$ into the formula, and simplify to find the value of the annuity after 6 years.

\[{S}_{72}=\frac{50(1-{1.005}^{72})}{1-1.005}\approx 4\text{,}320.44\]

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 $\text{72(50) = \$3,600}\text{.}$ This means that because of the annuity, the parent earned $720.44 interest in their college fund.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

sum of the first $n$ terms of an arithmetic series	${S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}$
sum of the first $n$ terms of a geometric series	${S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r},r\ne 1$
sum of an infinite geometric series with \(-1	${S}_{n}=\frac{{a}_{1}}{1-r},r\ne 1$

Key Concepts

The sum of the terms in a sequence is called a series.
A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum. See .
The sum of the terms in an arithmetic sequence is called an arithmetic series.
The sum of the first $n$ terms of an arithmetic series can be found using a formula. See and .
The sum of the terms in a geometric sequence is called a geometric series.
The sum of the first $n$ terms of a geometric series can be found using a formula. See and .
The sum of an infinite series exists if the series is geometric with \(-1
If the sum of an infinite series exists, it can be found using a formula. See , , and .
An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series. See .

Series and Their Notations

Use summation notation to write a sum. (IA 12.1.5)
Find the sum of the first n terms of an arithmetic sequence. (IA 12.2.3)

A series is the sum of the terms of a sequence. For example, 1 + 6 + 11+ 16 + 21 + 26 + 31 is the sum of the first seven terms arithmetic sequence with general term, ${a}_{n}=5n-4.$

We write a series by using the summation notation. In order to write that summation, we will need to find the general term of our sequence and the summation will look like:

For the series, 1 + 6 + 11 + 16 + 21 + 26 + 31 + .... the summation notation is $\sum _{n=1}^{7}5n-4$

Example

Use summation notation to write the sum.

Try it.

Write the sum using summation notation: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$

Solution

	$\begin{array}{llllll}n: & 1, & 2, & 3, & 4, & 5\end{array}$
Look for a pattern in the terms.	Terms: $\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}$
The numerators are all one. The denominators are the counting numbers from 1 thru 5.	Pattern: $\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}$
The general terms is:	$\frac{1}{n}$
The sum in summation notation is:	$\sum _{n=1}^{5}\frac{1}{n}$

Use summation notation to write the sum.

Try it.

Write the sum using summation notation: $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$

Try it.

Write the sum using summation notation: $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}$

Try it.

Find the sum of the first 30 terms of the arithmetic sequence: 16, 10, 4, –2, –8, ………

Try it.

Find the sum of the first 50 terms of the arithmetic sequence whose general term is ${a}_{n}=2n+7$ .

Try it.

Find the sum: $\sum _{i=1}^{30}(7i-5)$

Condensed — the full section is in OpenStax Precalculus 2e.

Using Summation Notation

\[3+7+11+15+19+...\]

The nth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation ${S}_{n}$ represents the partial sum.

\[\begin{array}{l}{S}_{1}=3 \\ {S}_{2}=3+7=10 \\ {S}_{3}=3+7+11=21 \\ {S}_{4}=3+7+11+15=36\end{array}\]

\[\begin{array}{l}\begin{array}{l} \\ {a}_{1}=2(1)=2\end{array} \\ {a}_{2}=2(2)=4 \\ {a}_{3}=2(3)=6 \\ {a}_{4}=2(4)=8 \\ {a}_{5}=2(5)=10\end{array}\]

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

\[\sum _{k=1}^{5}2k=2+4+6+8+10=30\]

Condensed — the full section is in OpenStax Precalculus 2e.

Using the Formula for Arithmetic Series

\[{S}_{n}={a}_{1}+({a}_{1}+d)+({a}_{1}+2d)+...+({a}_{n}-d)+{a}_{n}.\]

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

\[{S}_{n}={a}_{n}+({a}_{n}-d)+({a}_{n}-2d)+...+({a}_{1}+d)+{a}_{1}.\]

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

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

\[2{S}_{n}=n({a}_{1}+{a}_{n}).\]

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

\[{S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\]

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

Condensed — the full section is in OpenStax Precalculus 2e.

Using the Formula for Geometric Series

\[{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n-1}{a}_{1}.\]

\[r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\]

Next, we subtract this equation from the original equation.

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 ${S}_{n},$ divide both sides by $(1-r).$

\[{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r}\ \text{r}\ne \text{1}\]

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

Example

Try it.

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

\[\begin{array}{l}{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r} \\ {S}_{5}=\frac{26\text{,}750(1-{1.016}^{5})}{1-1.016}\approx 138\text{,}099.03\end{array}\]

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

Condensed — the full section is in OpenStax Precalculus 2e.

Using the Formula for the Sum of an Infinite Geometric Series

Condensed — the full section is in OpenStax Precalculus 2e.

Solving Annuity Problems

\[{S}_{72}=\frac{50(1-{1.005}^{72})}{1-1.005}\approx 4\text{,}320.44\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

sum of the first $n$ terms of an arithmetic series	${S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}$
sum of the first $n$ terms of a geometric series	${S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r},r\ne 1$
sum of an infinite geometric series with \(-1	${S}_{n}=\frac{{a}_{1}}{1-r},r\ne 1$

Key Concepts

The sum of the terms in a sequence is called a series.
A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum. See .
The sum of the terms in an arithmetic sequence is called an arithmetic series.
The sum of the first $n$ terms of an arithmetic series can be found using a formula. See and .
The sum of the terms in a geometric sequence is called a geometric series.
The sum of the first $n$ terms of a geometric series can be found using a formula. See and .
The sum of an infinite series exists if the series is geometric with \(-1
If the sum of an infinite series exists, it can be found using a formula. See , , and .
An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Write the sum using summation notation: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$

បង្ហាញចម្លើយ

	$\begin{array}{llllll}n: & 1, & 2, & 3, & 4, & 5\end{array}$
Look for a pattern in the terms.	Terms: $\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}$
The numerators are all one. The denominators are the counting numbers from 1 thru 5.	Pattern: $\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}$
The general terms is:	$\frac{1}{n}$
The sum in summation notation is:	$\sum _{n=1}^{5}\frac{1}{n}$

Write the sum using summation notation: $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$
Write the sum using summation notation: $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}$

ⓐ Find the sum of the first 30 terms of the arithmetic sequence: 7, 10, 13, 13, 19,...
ⓑ Find the sum of the first 50 terms of the arithmetic sequence whose general term is ${a}_{n}=2n-5$ .

បង្ហាញចម្លើយ

ⓐ

To find the sum of the 30 first terms, we use the formula ${S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})$.

We know that ${a}_{1}=7,d=3,andn=30$ but we need to find ${a}_{30}$ .

To find the 30th term, use the formula ${a}_{1}=7,d=3andn=30$ .	${a}_{n}={a}_{1}+(n-1)d$
Substitute	${a}_{30}=7+(30-1)(3)$
Simplify	$\begin{array}{l}{a}_{30}=7+(29)(3) \\ {a}_{30}=7+87 \\ {a}_{30}=94\end{array}$
To find ${S}_{30}$ use the formula with ${a}_{1}=7,{a}_{30}=94andn=30.$	${S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})$
Substitute and simplify	$\begin{array}{l}{S}_{30}=\frac{30}{2}(7+94) \\ {S}_{30}=15(101) \\ {S}_{30}=1515\end{array}$

ⓑ

To the sum of the first 50 terms of the arithmetic sequence whose general term is ${a}_{n}=2n-5$ . We need to find ${a}_{1}\text{and}{a}_{50}$ and substitute in the formula.

Find ${a}_{1}$	$\begin{array}{l}{a}_{n}=2n-5 \\ {a}_{1}=2(1)-5 \\ {a}_{1}=-3\end{array}$
Find ${a}_{50}$	$\begin{array}{l}{a}_{n}=2n-5 \\ {a}_{50}=2(50)-5 \\ {a}_{50}=95\end{array}$
Then find ${S}_{50}$ , use the formula with ${a}_{1}=-3,{a}_{50}=95\text{and}n=50.$	${S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})$
Substitute and simplify	$\begin{array}{l}{S}_{50}=\frac{50}{2}(-3+95) \\ {S}_{50}=25(92)\end{array}$
	${S}_{50}=2300$

ⓒ

$\sum _{i=1}^{30}(6i-4)$ means find the sum of the first 30 terms of the sequence whose general term is $6i-4$ . We need to find ${a}_{1}\text{and}{a}_{30}$ and substitute in the formula.

Find ${a}_{1}$	$\begin{array}{l}{a}_{i}=6i-4 \\ {a}_{1}=6(1)-4 \\ {a}_{1}=2\end{array}$
Find ${a}_{30}$	$\begin{array}{l}{a}_{i}=6i-4 \\ {a}_{30}=6(30)-4 \\ {a}_{30}=176\end{array}$
Then find ${S}_{30}$ , use the formula with ${a}_{1}=2,{a}_{30}=176andn=30.$	${S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})$
Substitute and simplify	$\begin{array}{l}{S}_{30}=\frac{30}{2}(2+176) \\ {S}_{30}=15(178) \\ {S}_{30}=2630\end{array}$

Find the sum of the first 30 terms of the arithmetic sequence: 16, 10, 4, –2, –8, ………
Find the sum of the first 50 terms of the arithmetic sequence whose general term is ${a}_{n}=2n+7$ .
Find the sum: $\sum _{i=1}^{30}(7i-5)$
Evaluate $\sum _{k=3}^{7}{k}^{2}.$

បង្ហាញចម្លើយ

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 ${k}^{2}$ from $k=3$ to $k=7.$ We find the terms of the series by substituting $k=3\text{,}4\text{,}5\text{,}6\text{,}$ and $7$ into the function ${k}^{2}.$ We add the terms to find the sum.
\[\begin{array}{ll}\sum _{k=3}^{7}{k}^{2} & ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2} \\ & =9+16+25+36+49 \\ & =135\end{array}\]
Evaluate $\sum _{k=2}^{5}(3k-1).$

បង្ហាញចម្លើយ

38
Find the sum of each arithmetic series.
1. ⓐ $\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}$
2. ⓑ $\text{20 + 15 + 10 +\ldots + -50}$
3. ⓒ $\sum _{k=1}^{12}3k-8$
បង្ហាញចម្លើយ
1. ⓐ
  We are given ${a}_{1}=5$ and ${a}_{n}=32.$
  Count the number of terms in the sequence to find $n=10.$
  Substitute values for ${a}_{1},{a}_{n}\text{,}$ and $n$ into the formula and simplify.
  \[\begin{array}{l}\begin{array}{l} \\ {S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\end{array} \\ {S}_{10}=\frac{10(5+32)}{2}=185\end{array}\]
2. ⓑ
  We are given ${a}_{1}=20$ and ${a}_{n}=-50.$
  Use the formula for the general term of an arithmetic sequence to find $n.$
  \[\begin{array}{l}\ \ {a}_{n}={a}_{1}+(n-1)d \\ -50=20+(n-1)(-5) \\ -70=(n-1)(-5) \\ \ \ 14=n-1 \\ \ \ 15=n\end{array}\]
  Substitute values for ${a}_{1},{a}_{n}\text{,}\ n$ into the formula and simplify.
  \[\begin{array}{l}\begin{array}{l} \\ {S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\end{array} \\ {S}_{15}=\frac{15(20-50)}{2}=-225\end{array}\]
3. ⓒ
  To find ${a}_{1},$ substitute $k=1$ into the given explicit formula.
  \[\begin{array}{l}{a}_{k}=3k-8 \\ \ {a}_{1}=3(1)-8=-5\end{array}\]
  We are given that $n=12.$ To find ${a}_{12},$ substitute $k=12$ into the given explicit formula.
  \[\begin{array}{l}\ {a}_{k}=3k-8 \\ {a}_{12}=3(12)-8=28\end{array}\]
  Substitute values for ${a}_{1},{a}_{n},$ and $n$ into the formula and simplify.
  \[\begin{array}{l}\ {S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2} \\ {S}_{12}=\frac{12(-5+28)}{2}=138\end{array}\]
$\text{1}\text{.4 + 1}\text{.6 + 1}\text{.8 + 2}\text{.0 + 2}\text{.2 + 2}\text{.4 + 2}\text{.6 + 2}\text{.8 + 3}\text{.0 + 3}\text{.2 + 3}\text{.4}$

បង្ហាញចម្លើយ

$\text{26}\text{.4}$
$\text{13 + 21 + 29 + }\ldots \text{+ 69}$

បង្ហាញចម្លើយ

$\text{328}$
$\sum _{k=1}^{10}5-6k$

បង្ហាញចម្លើយ

$\text{-280}$
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?

បង្ហាញចម្លើយ

This problem can be modeled by an arithmetic series with ${a}_{1}=\frac{1}{2}$ and $d=\frac{1}{4}.$ We are looking for the total number of miles walked after 8 weeks, so we know that $n=8\text{,}$ and we are looking for ${S}_{8}.$ To find ${a}_{8},$ we can use the explicit formula for an arithmetic sequence.
\[\begin{array}{l}\begin{array}{l} \\ {a}_{n}={a}_{1}+d(n-1)\end{array} \\ {a}_{8}=\frac{1}{2}+\frac{1}{4}(8-1)=\frac{9}{4}\end{array}\]
We can now use the formula for arithmetic series.
\[\begin{array}{l}{S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2} \\ \ {S}_{8}=\frac{8(\frac{1}{2}+\frac{9}{4})}{2}=11\end{array}\]
She will have walked a total of 11 miles.
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?

បង្ហាញចម្លើយ

$2,025
Use the formula to find the indicated partial sum of each geometric series.
1. ⓐ ${S}_{11}$ for the series $\text{8 + -4 + 2 + }\ldots$
2. ⓑ ${\sum }^{\text{}}_{k=1}^{6}3⋅{2}^{k}$
បង្ហាញចម្លើយ
1. ⓐ
  ${a}_{1}=8,$ and we are given that $n=11.$
  We can find $r$ by dividing the second term of the series by the first.
  \[r=\frac{-4}{8}=-\frac{1}{2}\]
  Substitute values for ${a}_{1},r,\text{and}n$ into the formula and simplify.
  \[\begin{array}{l}{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r} \\ {S}_{11}=\frac{8(1-{(-\frac{1}{2})}^{11})}{1-(-\frac{1}{2})}\approx 5.336\end{array}\]
2. ⓑ
  Find ${a}_{1}$ by substituting $k=1$ into the given explicit formula.
  \[{a}_{1}=3⋅{2}^{1}=6\]
  We can see from the given explicit formula that $r=2.$ The upper limit of summation is 6, so $n=6.$
  Substitute values for ${a}_{1},\ r,$ and $n$ into the formula, and simplify.
  \[\begin{array}{l}{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r} \\ {S}_{6}=\frac{6(1-{2}^{6})}{1-2}=378\end{array}\]
${S}_{20}$ for the series $\text{1,000 + 500 + 250 + }\ldots$

បង្ហាញចម្លើយ

$\approx 2,000.00$
$\sum _{k=1}^{8}{3}^{k}$

បង្ហាញចម្លើយ

9,840
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.

បង្ហាញចម្លើយ

The problem can be represented by a geometric series with ${a}_{1}=26,750\text{;}$ $n=5\text{;}$ and $r=1.016.$ Substitute values for ${a}_{1}\text{,}$ $r\text{,}$ and $n$ into the formula and simplify to find the total amount earned at the end of 5 years.
\[\begin{array}{l}{S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r} \\ {S}_{5}=\frac{26\text{,}750(1-{1.016}^{5})}{1-1.016}\approx 138\text{,}099.03\end{array}\]
He will have earned a total of $138,099.03 by the end of 5 years.
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?

បង្ហាញចម្លើយ

$275,513.31
Determine whether the sum of each infinite series is defined.
1. ⓐ $\text{12 + 8 + 4 + }\ldots$
2. ⓑ $\frac{3}{4}+\frac{1}{2}+\frac{1}{3}+...$
3. ⓒ $\sum _{k=1}^{\infty }27⋅{(\frac{1}{3})}^{k}$
4. ⓓ $\sum _{k=1}^{\infty }5k$
បង្ហាញចម្លើយ
1. ⓐThe ratio of the second term to the first is $\frac{\text{2}}{\text{3}},$ which is not the same as the ratio of the third term to the second, $\frac{1}{2}.$ 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 $\frac{2}{3}\text{.}$ The sum of the infinite series is defined.
3. ⓒThe given formula is exponential with a base of $\frac{1}{3}\text{;}$ the series is geometric with a common ratio of $\frac{1}{3}\text{.}$ 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.
$\frac{1}{3}+\frac{1}{2}+\frac{3}{4}+\frac{9}{8}+...$

បង្ហាញចម្លើយ

The sum is not defined.
$24+(-12)+6+(-3)+...$

បង្ហាញចម្លើយ

The sum of the infinite series is defined.
$\sum _{k=1}^{\infty }15⋅{(-0.3)}^{k}$

បង្ហាញចម្លើយ

The sum of the infinite series is defined.
Find the sum, if it exists, for the following:
1. ⓐ $10+9+8+7+\ldots$
2. ⓑ $248.6+99.44+39.776+\ \ldots$
3. ⓒ $\sum _{k=1}^{\infty }4\text{,}374⋅{(-\frac{1}{3})}^{k-1}$
4. ⓓ $\sum _{k=1}^{\infty }\frac{1}{9}⋅{(\frac{4}{3})}^{k}$
បង្ហាញចម្លើយ
1. ⓐThere is not a constant ratio; the series is not geometric.
2. ⓑ
  There is a constant ratio; the series is geometric. ${a}_{1}=248.6$ and $r=\frac{99.44}{248.6}=0.4,$ so the sum exists. Substitute ${a}_{1}=248.6$ and $r=0.4$ into the formula and simplify to find the sum:
  \[\begin{array}{l}S=\frac{{a}_{1}}{1-r} \\ S=\frac{248.6}{1-0.4}=414.\overset{\bar}{3}\end{array}\]
3. ⓒ
  The formula is exponential, so the series is geometric with $r=-\frac{1}{3}.$ Find ${a}_{1}$ by substituting $k=1$ into the given explicit formula:
  \[{a}_{1}=4\text{,}374⋅{(-\frac{1}{3})}^{1-1}=4\text{,}374\]
  Substitute ${a}_{1}=4\text{,}374$ and $r=-\frac{1}{3}$ into the formula, and simplify to find the sum:
  \[\begin{array}{l}S=\frac{{a}_{1}}{1-r} \\ S=\frac{4\text{,}374}{1-(-\frac{1}{3})}=3\text{,}280.5\end{array}\]
4. ⓓThe formula is exponential, so the series is geometric, but $r>1.$ The sum does not exist.
Find an equivalent fraction for the repeating decimal $0.\overset{\bar}{3}$

បង្ហាញចម្លើយ

We notice the repeating decimal $0.\overset{\bar}{3}=0.333...$ so we can rewrite the repeating decimal as a sum of terms.
\[0.\overset{\bar}{3}=0.3+0.03+0.003+...\]
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
\[{S}_{n}=\frac{{a}_{1}}{1-r}=\frac{0.3}{1-0.1}=\frac{0.3}{0.9}=\frac{1}{3}.\]
$2+\frac{2}{3}+\frac{2}{9}+...$

បង្ហាញចម្លើយ

3
$\sum _{k=1}^{\infty }0.76k+1$

បង្ហាញចម្លើយ

The series is not geometric.
$\sum _{k=1}^{\infty }{(-\frac{3}{8})}^{k}$

បង្ហាញចម្លើយ

$-\frac{3}{11}$
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?

បង្ហាញចម្លើយ

The value of the initial deposit is $100, so ${a}_{1}=100.$ A total of 120 monthly deposits are made in the 10 years, so $n=120.$ To find $r,$ divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.
\[r=1+\frac{0.09}{12}=1.0075\]
Substitute ${a}_{1}=100\text{,}\ r=1.0075\text{,}\ \text{and}\ n=120$ into the formula for the sum of the first $n$ terms of a geometric series, and simplify to find the value of the annuity.
\[{S}_{120}=\frac{100(1-{1.0075}^{120})}{1-1.0075}\approx 19\text{,}351.43\]
So the account has $19,351.43 after the last deposit is made.
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?

បង្ហាញចម្លើយ

$32,775.87
What is an $n\text{th}$ partial sum?

បង្ហាញចម្លើយ

An $n\text{th}$ partial sum is the sum of the first $n$ terms of a sequence.
What is the difference between an arithmetic sequence and an arithmetic series?
What is a geometric series?

បង្ហាញចម្លើយ

A geometric series is the sum of the terms in a geometric sequence.
How is finding the sum of an infinite geometric series different from finding the $n\text{th}$ partial sum?
What is an annuity?

បង្ហាញចម្លើយ

An annuity is a series of regular equal payments that earn a constant compounded interest.
The sum of terms ${m}^{2}+3m$ from $m=1$ to $m=5$
The sum from of $n=0$ to $n=4$ of $5n$

បង្ហាញចម្លើយ

$\sum _{n=0}^{4}5n$
The sum of $6k-5$ from $k=-2$ to $k=1$
The sum that results from adding the number 4 five times

បង្ហាញចម្លើយ

$\sum _{k=1}^{5}4$

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Series and Their Notations

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.
Use summation notation to write a sum. (IA 12.1.5)
Find the sum of the first n terms of an arithmetic sequence. (IA 12.2.3)
Identify the lower limit of summation.

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

ព្យាយាមរបស់អ្នកផ្ទាល់

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

បន្ថែមទៀតក្នុង Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

	\(\begin{array}{llllll}n: & 1, & 2, & 3, & 4, & 5\end{array}\)
Look for a pattern in the terms.	Terms: \(\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}\)
The numerators are all one. The denominators are the counting numbers from 1 thru 5.	Pattern: \(\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}\)
The general terms is:	\(\frac{1}{n}\)
The sum in summation notation is:	\(\sum _{n=1}^{5}\frac{1}{n}\)

sum of the first \(n\) terms of an arithmetic series	\({S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\)
sum of the first \(n\) terms of a geometric series	\({S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r},r\ne 1\)
sum of an infinite geometric series with \(-1	\({S}_{n}=\frac{{a}_{1}}{1-r},r\ne 1\)

	\(\begin{array}{llllll}n: & 1, & 2, & 3, & 4, & 5\end{array}\)
Look for a pattern in the terms.	Terms: \(\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}\)
The numerators are all one. The denominators are the counting numbers from 1 thru 5.	Pattern: \(\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}\)
The general terms is:	\(\frac{1}{n}\)
The sum in summation notation is:	\(\sum _{n=1}^{5}\frac{1}{n}\)

sum of the first \(n\) terms of an arithmetic series	\({S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\)
sum of the first \(n\) terms of a geometric series	\({S}_{n}=\frac{{a}_{1}(1-{r}^{n})}{1-r},r\ne 1\)
sum of an infinite geometric series with \(-1	\({S}_{n}=\frac{{a}_{1}}{1-r},r\ne 1\)

	\(\begin{array}{llllll}n: & 1, & 2, & 3, & 4, & 5\end{array}\)
Look for a pattern in the terms.	Terms: \(\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}\)
The numerators are all one. The denominators are the counting numbers from 1 thru 5.	Pattern: \(\begin{array}{llllll} & 1, & \frac{1}{2}, & \frac{1}{3}, & \frac{1}{4}, & \frac{1}{5}\end{array}\)
The general terms is:	\(\frac{1}{n}\)
The sum in summation notation is:	\(\sum _{n=1}^{5}\frac{1}{n}\)

To find the 30th term, use the formula \({a}_{1}=7,d=3andn=30\) .	\({a}_{n}={a}_{1}+(n-1)d\)
Substitute	\({a}_{30}=7+(30-1)(3)\)
Simplify	\(\begin{array}{l}{a}_{30}=7+(29)(3) \\ {a}_{30}=7+87 \\ {a}_{30}=94\end{array}\)
To find \({S}_{30}\) use the formula with \({a}_{1}=7,{a}_{30}=94andn=30.\)	\({S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})\)
Substitute and simplify	\(\begin{array}{l}{S}_{30}=\frac{30}{2}(7+94) \\ {S}_{30}=15(101) \\ {S}_{30}=1515\end{array}\)

Find \({a}_{1}\)	\(\begin{array}{l}{a}_{n}=2n-5 \\ {a}_{1}=2(1)-5 \\ {a}_{1}=-3\end{array}\)
Find \({a}_{50}\)	\(\begin{array}{l}{a}_{n}=2n-5 \\ {a}_{50}=2(50)-5 \\ {a}_{50}=95\end{array}\)
Then find \({S}_{50}\) , use the formula with \({a}_{1}=-3,{a}_{50}=95\text{and}n=50.\)	\({S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})\)
Substitute and simplify	\(\begin{array}{l}{S}_{50}=\frac{50}{2}(-3+95) \\ {S}_{50}=25(92)\end{array}\)
	\({S}_{50}=2300\)

Find \({a}_{1}\)	\(\begin{array}{l}{a}_{i}=6i-4 \\ {a}_{1}=6(1)-4 \\ {a}_{1}=2\end{array}\)
Find \({a}_{30}\)	\(\begin{array}{l}{a}_{i}=6i-4 \\ {a}_{30}=6(30)-4 \\ {a}_{30}=176\end{array}\)
Then find \({S}_{30}\) , use the formula with \({a}_{1}=2,{a}_{30}=176andn=30.\)	\({S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})\)
Substitute and simplify	\(\begin{array}{l}{S}_{30}=\frac{30}{2}(2+176) \\ {S}_{30}=15(178) \\ {S}_{30}=2630\end{array}\)

Series and Their Notations

Using Summation Notation

Using the Formula for Arithmetic Series

Using the Formula for Geometric Series

Using the Formula for the Sum of an Infinite Geometric Series

Solving Annuity Problems

Key Equations

Key Concepts

Series and Their Notations

Using Summation Notation

Using the Formula for Arithmetic Series

Using the Formula for Geometric Series

Using the Formula for the Sum of an Infinite Geometric Series

Solving Annuity Problems

Key Equations

Key Concepts

Practice (40)

Symbols used here

How to: Series and Their Notations

Questions people ask

ព្យាយាម​របស់​អ្នក​ផ្ទាល់

បន្ថែម​ទៀត​ក្នុង Precalculus

ព្យាយាមរបស់អ្នកផ្ទាល់

បន្ថែមទៀតក្នុង Precalculus