13. Sequences, Series, and the Binomial Theorem

Introduction to Sequences

13. Sequences, Series, and the Binomial Theorem

Introduction to Sequences: Videos & Practice Problems

Topic summary

Sequences are functions with domains of positive integers, where each term $a_{n}$ is found by substituting the index $n$ . They can be finite or infinite, showing patterns like arithmetic progressions with constant differences or exponential expressions such as $a_{n} =2 n$ . Recognizing patterns, including alternating signs using $(-1)^n$ , and fractional terms helps derive explicit formulas. Understanding these concepts aids in modeling numerical sequences and polynomial terms, emphasizing the role of exponents, coefficients, and powers in standard form expressions.

concept

Intro to Sequences

Video duration:

Intro to Sequences Video Summary

Sequences are fundamental mathematical concepts used to model and predict patterns both in theoretical contexts and real-world applications. Essentially, a sequence is a function where the domain consists of positive integers starting from one, two, three, and so forth. Each input, called an index (denoted by n), corresponds to an output known as a term, represented as a_n. This notation allows us to express sequences as ordered lists of numbers, where a₁ is the first term, a₂ the second, and so on.

Understanding sequences involves evaluating the general term formula by substituting different index values. For example, if the general term is given by a_n = 2ⁿ, the first term is found by plugging in n = 1, yielding a₁ = 2¹ = 2. Similarly, the second term is a₂ = 2² = 4, and the third term is a₃ = 2³ = 8. This process can be continued to find any term in the sequence, such as the tenth term, calculated as a₁₀ = 2¹⁰ = 1024. Representing sequences as lists, such as 2, 4, 8, ..., helps visualize the pattern and understand its progression.

Sequences can be classified as either infinite or finite. An infinite sequence continues indefinitely without end, as seen in the example above where the index n can take any positive integer value. In contrast, a finite sequence has a restricted domain, such as when n is limited to values between 1 and 5, resulting in a sequence that stops after a certain number of terms. Recognizing these distinctions is crucial for analyzing and working with sequences in various mathematical contexts.

Study Smarter with Worksheets.

Follow along with each video using our printable worksheets

Problem

List the first four terms of each sequence whose general term is given.
(A) $a_{n} = 5 - 2 n$

$5 comma 3 comma 1 comma negative 1$

$3, 1, - 1, - 3$

$3 comma 1 comma 0 comma negative 1$

$5 comma 3 comma 1 comma 0$

Problem

List the first four terms of each sequence whose general term is given.
(B) $a_{n} = n^{2} - 4 n + 2$

$- 1, - 2, - 1, 2$

$2, - 1, - 2, - 1$

$0, - 1, - 2, - 1$

$- 2, - 1, 2, 7$

example

Intro to Sequences Example 1

Video duration:

Intro to Sequences Example 1 Video Summary

Alternating sequences are characterized by terms that switch signs between positive and negative as the sequence progresses. This behavior is often generated by including a factor of negative one raised to a power involving the term index n. For example, consider the sequence defined by the general term $a_n = (-1)^n \times 2^{n-1}$. Calculating the first four terms, we start with $a_1 = (-1)^1 \times 2^{0} = -1 \times 1 = -1$. The second term is $a_2 = (-1)^2 \times 2^{1} = 1 \times 2 = 2$. The third term, $a_3 = (-1)^3 \times 2^{2} = -1 \times 4 = -4$, and the fourth term, $a_4 = (-1)^4 \times 2^{3} = 1 \times 8 = 8$. This sequence alternates signs: negative, positive, negative, positive, illustrating the defining feature of an alternating sequence.

Another example is the sequence given by $a_n = (-1)^{n+1} \times n^2$. Here, the exponent on $-1$ is shifted by one, which affects the starting sign. The first term is $a_1 = (-1)^{1+1} \times 1^2 = (-1)^2 \times 1 = 1$. The second term is $a_2 = (-1)^{2+1} \times 2^2 = (-1)^3 \times 4 = -4$. The third term is $a_3 = (-1)^{3+1} \times 3^2 = (-1)^4 \times 9 = 9$, and the fourth term is $a_4 = (-1)^{4+1} \times 4^2 = (-1)^5 \times 16 = -16$. This sequence also alternates signs, starting positive and then switching to negative, positive, and negative again.

The key to understanding alternating sequences lies in the term $(-1)^n$ or $(-1)^{n+1}$. Since $-1$ raised to an even power is positive and to an odd power is negative, this factor causes the sequence terms to alternate in sign. When the exponent is simply $n$, the sequence begins with a negative term because $(-1)^1 = -1$. When the exponent is $n+1$, the sequence starts positive because $(-1)^2 = 1$. This alternating sign pattern is fundamental in many mathematical contexts, including series convergence and oscillatory behavior.

Problem

What is the value of the term being asked for?
$a_{n} = \frac{2 n + 1}{n + 4}$
$a sub 4 equals question mark$

$\frac{11}{9}$

$\frac{5}{6}$

$1$

$\frac{9}{8}$

concept

Intro to Sequences Example 2

Video duration:

Intro to Sequences Example 2 Video Summary

A finite sequence is a list of numbers defined by a specific rule and has a limited number of terms. In this example, the sequence is given by the formula $a_n = 4 - n$ for values of $n$ ranging from 1 to 6. To find each term, substitute the values of $n$ into the formula:

For $n = 1$, the first term is $a_1 = 4 - 1 = 3$.
For $n = 2$, the second term is $a_2 = 4 - 2 = 2$.
For $n = 3$, the third term is $a_3 = 4 - 3 = 1$.
For $n = 4$, the fourth term is $a_4 = 4 - 4 = 0$.
For $n = 5$, the fifth term is $a_5 = 4 - 5 = -1$.
For $n = 6$, the sixth and final term is $a_6 = 4 - 6 = -2$.

Thus, the finite sequence consists of the terms: \$3, 2, 1, 0, -1, -2$. Unlike infinite sequences, which continue indefinitely and are often represented with ellipses (...), finite sequences have a clear endpoint and list all their terms explicitly. Understanding how to evaluate each term using the given formula is essential for working with sequences in algebra and calculus.

example

Intro to Sequences Example 3

Video duration:

Intro to Sequences Example 3 Video Summary

The finite sequence defined by the formula a_n = n / (2n + 1) for n = 1, 2, 3, 4, 5 consists of five terms. To find each term, substitute the value of n into the formula and simplify the fraction.

For the first term, when n = 1, the term is calculated as:

\[a_1 = \frac{1}{2 \times 1 + 1} = \frac{1}{3}\]

The second term, with n = 2, is:

\[a_2 = \frac{2}{2 \times 2 + 1} = \frac{2}{5}\]

For the third term, n = 3, the value is:

\[a_3 = \frac{3}{2 \times 3 + 1} = \frac{3}{7}\]

The fourth term, when n = 4, is:

\[a_4 = \frac{4}{2 \times 4 + 1} = \frac{4}{9}\]

Finally, the fifth term, with n = 5, is:

\[a_5 = \frac{5}{2 \times 5 + 1} = \frac{5}{11}\]

Thus, the sequence is explicitly given by the list of terms:

\[\left\{ \frac{1}{3}, \frac{2}{5}, \frac{3}{7}, \frac{4}{9}, \frac{5}{11} \right\}\]

This sequence is finite because it is defined only for the integer values of n from 1 to 5, and it does not continue beyond these terms. Understanding how to evaluate each term by substituting values into the formula helps in recognizing the pattern and behavior of finite sequences in general.

example

Intro to Sequences Example 4

Video duration:

Intro to Sequences Example 4 Video Summary

When analyzing sequences to find a specific term, identifying the underlying pattern is essential. For example, consider the sequence 2, 5, 8, 11, 14. Observing the differences between consecutive terms reveals a constant increase of 3. This indicates an arithmetic sequence where each term increases by a common difference of 3. Starting from the first term, 2, the sequence progresses by adding 3 repeatedly. To find the tenth term, continue this pattern: the sixth term is 17, the seventh is 20, the eighth is 23, the ninth is 26, and the tenth term is 29. Thus, the tenth term, denoted as $a_{10}$, equals 29.

In another example, consider the sequence of fractions: $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{4}$, $\frac{4}{5}$, $\frac{5}{6}$. Here, both the numerator and denominator increase by 1 with each successive term. This pattern can be expressed as the $n$th term having a numerator of $n$ and a denominator of $n+1$, or $a_n = \frac{n}{n+1}$. To find the twelfth term, continue the pattern up to $n=12$, resulting in $\frac{12}{13}$. Therefore, the twelfth term, $a_{12}$, is $\frac{12}{13}$.

Recognizing these patterns allows for the determination of specific terms in a sequence without listing all preceding terms. Arithmetic sequences are characterized by a constant difference between terms, while sequences like the fractional example can often be described by a formula involving the term number $n$. Understanding these patterns and their general formulas is fundamental in sequence analysis and helps efficiently find any term in the sequence.

concept

Writing a General Formula

Video duration:

Writing a General Formula Video Summary

In the study of sequences, understanding how to derive explicit formulas is essential. An explicit formula allows you to calculate the terms of a sequence based on their position, denoted by $ n $. The process often involves identifying patterns within the sequence of numbers.

One common pattern is an arithmetic sequence, where the terms increase by a constant amount. For example, in the sequence 5, 6, 7, 8, 9, each term increases by 1. The general formula for such sequences can be expressed as $ a_n = n + c $, where $ c $ is a constant that adjusts the starting point. In this case, since the first term is 5, we find $ c = 4 $, leading to the formula $ a_n = n + 4 $. This can be verified by substituting values of $ n $ to ensure the correct terms are produced.

Another scenario involves alternating sequences, such as -5, 5, -5, 5. Here, the sign alternates, which can be represented using $ (-1)^n $. To adjust for the specific values, we multiply by 5, resulting in the formula $ a_n = (-1)^n \cdot 5 $. This formula captures the alternating nature and the magnitude of the terms.

Sequences can also include fractions, as seen in 1/2, 2/3, 3/4, 4/5, 5/6. The numerators follow a simple pattern of increasing by 1, which corresponds directly to $ n $. The denominators also increase by 1 but start from 2, leading to the formula $ a_n = \frac{n}{n + 1} $. This formula accurately reflects the structure of the sequence.

Exponential sequences, such as 2, 4, 8, 16, 32, exhibit a different pattern where each term is a power of 2. The general formula for this type of sequence is $ a_n = 2^n $. This reflects the rapid growth of the terms, and substituting values of $ n $ confirms the correctness of the formula.

In summary, identifying the type of sequence and its underlying pattern is crucial for deriving the correct explicit formula. By analyzing the differences between terms, their signs, and their relationships, one can construct formulas that accurately represent the sequences. Practice with various examples will enhance your ability to recognize these patterns and formulate the corresponding equations.

example

Writing a General Formula Example 5

Video duration:

Writing a General Formula Example 5 Video Summary

In this example, we explore how to derive a general formula for a sequence based on its first four terms. The sequence provided is:

1/1×2, 1/2×3, 1/3×4, 1/4×5.

To find the general term, denoted as $ a_n $, we first observe the pattern in the sequence. Notably, each term is a fraction, indicating that our general formula will also be a fraction. The numerators of all terms are consistently 1, which simplifies our formula since the numerator does not depend on $ n $. Thus, the numerator remains 1.

Next, we analyze the denominators, which are products of two numbers: for the first term, it is $ 1 \times 2 $; for the second, $ 2 \times 3 $; for the third, $ 3 \times 4 $; and for the fourth, $ 4 \times 5 $. The first number in each denominator corresponds directly to $ n $, as it increases by 1 with each term. Therefore, the first part of the denominator can be expressed as $ n $.

For the second number in the denominator, we see it also increases by 1: it is 2 for the first term, 3 for the second, 4 for the third, and 5 for the fourth. This indicates that the second number can be represented as $ n + 1 $ since it is always one more than $ n $.

Combining these observations, we can express the general formula for the sequence as:

\[ a_n = \frac{1}{n(n + 1)} \]

To find the 15th term of the sequence, we substitute $ n = 15 $ into our formula:

\[ a_{15} = \frac{1}{15(15 + 1)} = \frac{1}{15 \times 16} \]

Calculating this gives:

\[ a_{15} = \frac{1}{240} \approx 0.0041667 \] (repeating).

This process illustrates how to derive a general formula from a sequence and use it to calculate specific terms efficiently, avoiding the need to list all preceding terms.

example

Writing a General Formula Example 6

Video duration:

Writing a General Formula Example 6 Video Summary

In this exploration of sequences, we analyze a specific pattern characterized by the first five terms: -2, 4, -6, 8, -10. The goal is to derive a general formula for the nth term and subsequently calculate the 18th term. Recognizing the sequence's structure is crucial, as it alternates in sign while increasing in absolute value by 2 with each term.

To formulate the nth term, we first note the alternating signs, which can be represented mathematically as (-1)^n. Since the first term is negative, we use (-1)^n rather than (-1)^{n+1}. Next, to account for the consistent increase of 2, we multiply by 2n. Thus, the general formula for the nth term can be expressed as:

a_n = (-1)^n \cdot 2n

To verify this formula, we can substitute values for n:

For n = 1: a_1 = (-1)^1 \cdot 2 \cdot 1 = -2
For n = 2: a_2 = (-1)^2 \cdot 2 \cdot 2 = 4
For n = 3: a_3 = (-1)^3 \cdot 2 \cdot 3 = -6

These calculations confirm that the formula accurately represents the sequence. To find the 18th term, we substitute n = 18 into the formula:

a_{18} = (-1)^{18} \cdot 2 \cdot 18

Since (-1)^{18} = 1 (as 18 is an even number), we simplify this to:

a_{18} = 1 \cdot 36 = 36

Thus, the 18th term of the sequence is 36. This method of deriving a general formula not only streamlines the process of finding specific terms but also enhances understanding of the underlying patterns in sequences.

example

Writing a General Formula Example 7

Video duration:

Writing a General Formula Example 7 Video Summary

Carla saves money each week by increasing her savings by \$10 compared to the previous week. She starts by saving \$50 in the first week, then \$60 in the second week, \$70 in the third week, and so on. To find how much she will save in the eighth week, we can analyze the pattern or use a general formula for the sequence.

Observing the pattern, the savings increase by \$10 every week. Starting from \$50 in week one, the amounts saved are \$50, \$60, \$70, \$80, \$90, \$100, \$110, and finally \$120 in week eight. This shows a clear arithmetic sequence where the common difference is \$10.

To express this algebraically, the savings in week n can be represented by the formula for the nth term of an arithmetic sequence:

\[a_n = a_1 + (n - 1)d\]

where $a_1$ is the first term and $d$ is the common difference. Here, $a_1 = 50$ and $d = 10$. Substituting these values, the formula becomes:

\[a_n = 50 + 10(n - 1)\]

To find the savings in week eight, substitute $n = 8$:

\[a_8 = 50 + 10(8 - 1) = 50 + 10 \times 7 = 50 + 70 = 120\]

Therefore, Carla will save \$120 in the eighth week. This approach highlights the usefulness of arithmetic sequences in modeling real-life situations involving consistent incremental changes. Understanding how to derive and apply the general term formula allows for efficient calculation of any term in the sequence without enumerating all previous values.

Here’s what students ask on this topic:

A sequence in mathematics is a list of numbers arranged in a specific order, where each number is called a term. Sequences are closely related to functions because a sequence can be viewed as a function whose domain is the set of positive integers (1, 2, 3, ...). Each input, called the index $n$ , corresponds to a term in the sequence, denoted as $a_{n}$ . The general term or explicit formula of a sequence is a function that, when you substitute the index $n$ , gives the value of the term at that position. This connection allows us to use function concepts to analyze and predict patterns in sequences.

To find the terms of a sequence from its general term formula, you substitute the index $n$ with positive integers starting from 1. For example, if the general term is $a_{n} = 2 2^{n}$ , then the first term is found by plugging in $n = 1$ , giving $2^1 = 2$ . The second term is $2^2 = 4$ , and the third term is $2^3 = 8$ . You continue this process for as many terms as needed. This method works for both finite sequences, which have a limited number of terms, and infinite sequences, which continue indefinitely.

A finite sequence has a limited number of terms, meaning it stops after a certain index value. For example, if the domain of the sequence is restricted to $1 \leq n \leq 6$ , then the sequence has exactly six terms. An infinite sequence, on the other hand, continues indefinitely without stopping, with the domain being all positive integers $n \geq 1$ . Understanding whether a sequence is finite or infinite is important for determining how many terms to consider and how to analyze the sequence's behavior.

To find the explicit formula of a sequence from its terms, first identify the pattern in the sequence. If the terms increase by a constant difference, the sequence is arithmetic, and the formula will be linear, often of the form $a_{n} = dn + c$ , where $d$ is the common difference and $c$ is a constant. If the terms multiply by a constant factor, the sequence is geometric, and the formula involves an exponent, such as $a_{n} = r b^{n}$ . For alternating signs, include $(-1)^n$ to represent sign changes. By testing initial terms and adjusting constants, you can derive the explicit formula that generates the sequence.

Common patterns in sequences include arithmetic progressions where terms increase by a constant difference, geometric progressions where terms multiply by a constant ratio, alternating signs, and fractional patterns. In arithmetic sequences, the general term formula is linear, such as $a_{n} = dn + c$ . For geometric sequences, the formula involves exponents, like $a_{n} = r b^{n}$ . Alternating signs are modeled using $(-1)^n$ or $(-1)^{n+1}$ . Fractional sequences often have numerators and denominators following separate patterns, requiring formulas like $= \frac{n}{n+1}$ . Recognizing these patterns helps in writing accurate explicit formulas.

Your Intermediate Algebra tutors

Patrick Ford

Physics and Math Lead Instructor

Introduction to Sequences: Videos & Practice Problems