15. Sequences, Series, and the Binomial Theorem

Introduction to Sequences

15. 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 alternating signs modeled by ${(-1)}^{n}$ . Exponential sequences involve terms like $2^{n}$ . Understanding these patterns helps derive explicit formulas, essential for analyzing polynomial terms, exponents, and powers in standard form expressions.

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Intro to Sequences

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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, which is a formula defining the nth term of the sequence. For example, if the general term is given by a_n = 2ⁿ, finding specific terms requires substituting the index n with positive integers. Calculating the first three terms involves computing a₁ = 2¹ = 2, a₂ = 2² = 4, and a₃ = 2³ = 8. This sequence can be represented as the list 2, 4, 8, ..., where the ellipsis indicates the pattern continues indefinitely. To find the tenth term, substitute n = 10 into the general term, yielding a₁₀ = 2¹⁰ = 1024.

Sequences can be classified as either infinite or finite. An infinite sequence continues without end, as in the example above, where the index n can take any positive integer value. Conversely, a finite sequence has a restricted domain, such as when n is limited to values between 1 and 5, resulting in a sequence with only five terms. Recognizing these distinctions is crucial for analyzing and applying sequences effectively.

By viewing sequences as functions with discrete domains, one can leverage prior knowledge of functions to analyze patterns, predict future terms, and understand the behavior of numerical sequences. This foundational understanding paves the way for exploring more complex sequences and their applications in various mathematical and scientific fields.

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Problem

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

$5,3,1,-1$

$3,1,-1,-3$

$3,1,0,-1$

$5,3,1,0$

Problem

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

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

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

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

$-2,-1,2,7$

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Intro to Sequences Example 1

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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:

$a_1 = (-1)^1 \times 2^{1-1} = -1 \times 2^0 = -1 \times 1 = -1$

$a_2 = (-1)^2 \times 2^{2-1} = 1 \times 2^1 = 2$

$a_3 = (-1)^3 \times 2^{3-1} = -1 \times 2^2 = -4$

$a_4 = (-1)^4 \times 2^{4-1} = 1 \times 2^3 = 8$

These terms alternate in sign: negative, positive, negative, positive, illustrating the alternating nature of the sequence. The factor $(-1)^n$ causes this sign change because raising $-1$ to an odd power yields $-1$, and to an even power yields $+1$.

Another example is the sequence defined by $a_n = (-1)^{n+1} \times n^2$. The first four terms are:

$a_1 = (-1)^{1+1} \times 1^2 = (-1)^2 \times 1 = 1$

$a_2 = (-1)^{2+1} \times 2^2 = (-1)^3 \times 4 = -4$

$a_3 = (-1)^{3+1} \times 3^2 = (-1)^4 \times 9 = 9$

$a_4 = (-1)^{4+1} \times 4^2 = (-1)^5 \times 16 = -16$

Here, the signs alternate starting with a positive term because $(-1)^{n+1}$ evaluates to $+1$ when $n=1$ (since \$1+1=2\( is even), and then alternates with each subsequent term. This pattern of alternating signs is a hallmark of alternating sequences.

In general, the presence of the factor \)(-1)^n$ or $(-1)^{n+1}$ in the general term of a sequence ensures that the sequence alternates signs. When $n$ is odd, $(-1)^n = -1$, and when $n$ is even, $(-1)^n = 1$. Similarly, $(-1)^{n+1}$ shifts the starting sign depending on the parity of $n+1$. Understanding this mechanism allows for predicting the sign pattern of terms in alternating sequences, which is essential in analyzing their behavior and convergence in advanced mathematical contexts.

Problem

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

$\frac{11}{9}$

$\frac{5}{6}$

$1$

$\frac{9}{8}$

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Intro to Sequences Example 2

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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 defined by the formula a_n = 4 − n, where n ranges from 1 to 6. To find each term, substitute the values of n into the formula:

For n = 1, the first term is a₁ = 4 − 1 = 3.

For n = 2, the second term is a₂ = 4 − 2 = 2.

For n = 3, the third term is a₃ = 4 − 3 = 1.

For n = 4, the fourth term is a₄ = 4 − 4 = 0.

For n = 5, the fifth term is a₅ = 4 − 5 = −1.

For n = 6, the sixth term is a₆ = 4 − 6 = −2.

Thus, the finite sequence consists of the terms: 3, 2, 1, 0, −1, and −2. Unlike infinite sequences, finite sequences have a definite end, so they do not continue indefinitely. Understanding how to evaluate each term by substituting values into the sequence formula is essential for working with sequences in algebra and calculus.

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Intro to Sequences Example 3

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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 finite sequence is the list of these five terms: 1/3, 2/5, 3/7, 4/9, 5/11. Since this sequence is finite, it does not continue indefinitely, and no ellipsis is needed. Understanding how to evaluate each term by substituting values into the sequence formula is essential for working with finite sequences and recognizing their limits.

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Intro to Sequences Example 4

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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 consistent 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 suggests that the $n$th term has a numerator of $n$ and a denominator of $n+1$. Extending this pattern to find the twelfth term involves continuing to add 1 to both numerator and denominator. The sixth term is $\frac{6}{7}$, the seventh is $\frac{7}{8}$, and so on, until the twelfth term, $a_{12}$, which is $\frac{12}{13}$.

Recognizing these patterns allows for determining specific terms in sequences without explicitly knowing a general formula. However, arithmetic sequences can also be described by a general term formula: $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference. For the first sequence, $a_1 = 2$ and $d = 3$, so the tenth term can be calculated as $a_{10} = 2 + (10-1) \times 3 = 29$. Similarly, for sequences involving fractions with numerators and denominators increasing linearly, the general term can be expressed as $a_n = \frac{n}{n+1}$. This formula directly provides the $n$th term without enumerating all previous terms.

Understanding how to identify and extend patterns in sequences is fundamental in mathematics, enabling efficient calculation of terms and deeper comprehension of sequence behavior. Mastery of these concepts lays the groundwork for exploring more advanced topics such as arithmetic and geometric sequences, series, and their applications.

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Writing a General Formula

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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.

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Writing a General Formula Example 5

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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.

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Writing a General Formula Example 6

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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.

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Writing a General Formula Example 7

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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 algebra is a list of numbers arranged in a specific order, where each number is called a term. Mathematically, a sequence can be viewed as a function whose domain is the set of positive integers (1, 2, 3, ...). Each input value, 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 an expression involving $n$ that allows you to find any term by substituting the index. This connection to functions helps us analyze sequences using familiar function concepts, such as evaluating terms by plugging in values of $n$ .

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 so on. This process can be repeated to find as many terms as needed. For finite sequences, you stop at the given maximum index. For infinite sequences, the terms continue indefinitely.

Finite sequences have a limited number of terms, meaning their domain of indices $n$ is restricted to a specific range, such as $1 ≤ n ≤ 6$ . For example, the sequence defined by $a_{n} = 4 - n$ for $n = 1, 2, ..., 6$ is finite and has exactly six terms. Infinite sequences, on the other hand, continue indefinitely with no upper bound on $n$ . An example is $a_{n} = 2^n$ for all positive integers $n$ . Understanding whether a sequence is finite or infinite is important for analyzing its behavior and applications.

Sequences that increase by a constant difference are called arithmetic sequences. Their explicit formula has the form $a_{n} = d n + c$ , where $d$ is the common difference and $c$ is a constant. For example, the sequence 5, 6, 7, 8, 9 increases by 1 each time, so the formula is $a_{n} = n + 4$ . This is because plugging in $n = 1$ gives 5, the first term. Adjusting the formula by adding or subtracting constants ensures the sequence matches the given terms.

Sequences with alternating signs can be represented using the factor $(-1)^n$ or $(-1)^{n+1}$ . This factor alternates between -1 and 1 as $n$ increases, causing the terms to switch signs. For example, the sequence -5, 5, -5, 5, ... can be written as $a_{n} = (-1)^n imes 5$ . Multiplying by 5 scales the alternating signs to the correct magnitude. This method is useful for modeling sequences where the sign changes regularly.

An exponential sequence has terms that grow by powers of a fixed base. Its general form is $a_{n} = b^n$ , where $b$ is a constant base and $n$ is the term index. For example, the sequence 2, 4, 8, 16, 32 can be expressed as $a_{n} = 2^n$ . To find terms, substitute values of $n$ starting from 1. This type of sequence grows faster than arithmetic sequences and is common in modeling exponential growth or decay.

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Introduction to Sequences: Videos & Practice Problems

Intro to Sequences

Intro to Sequences Video Summary

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

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

Intro to Sequences Example 1

Intro to Sequences Example 1 Video Summary

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

Intro to Sequences Example 2

Intro to Sequences Example 2 Video Summary

Intro to Sequences Example 3

Intro to Sequences Example 3 Video Summary

Intro to Sequences Example 4

Intro to Sequences Example 4 Video Summary

Writing a General Formula

Writing a General Formula Video Summary

Writing a General Formula Example 5

Writing a General Formula Example 5 Video Summary

Writing a General Formula Example 6

Writing a General Formula Example 6 Video Summary

Writing a General Formula Example 7

Writing a General Formula Example 7 Video Summary

Here’s what students ask on this topic:

What is the definition of a sequence in algebra and how is it related to functions?

How do you find the terms of a sequence given its general term formula?

What are the differences between finite and infinite sequences?

How can you write the explicit formula for a sequence that increases by a constant difference?

How do you represent sequences with alternating signs using explicit formulas?

What is the general form of an exponential sequence and how do you find its terms?

Your Beginning & Intermediate Algebra tutors

Introduction to Sequences: Videos & Practice Problems

Intro to Sequences

Intro to Sequences Video Summary

List the first four terms of each sequence whose general term is given.(A) an=5−2na_n=5-2n

List the first four terms of each sequence whose general term is given.(B) an=n2−4n+2a_n=n^2-4n+2

Intro to Sequences Example 1

Intro to Sequences Example 1 Video Summary

What is the value of the term being asked for?an=2n+1n+4a_{n}=\frac{2n+1}{n+4}an=?a_{n}=?

Intro to Sequences Example 2

Intro to Sequences Example 2 Video Summary

Intro to Sequences Example 3

Intro to Sequences Example 3 Video Summary

Intro to Sequences Example 4

Intro to Sequences Example 4 Video Summary

Writing a General Formula

Writing a General Formula Video Summary

Writing a General Formula Example 5

Writing a General Formula Example 5 Video Summary

Writing a General Formula Example 6

Writing a General Formula Example 6 Video Summary

Writing a General Formula Example 7

Writing a General Formula Example 7 Video Summary

Here’s what students ask on this topic:

What is the definition of a sequence in algebra and how is it related to functions?

What is the definition of a sequence in algebra and how is it related to functions?

How do you find the terms of a sequence given its general term formula?

How do you find the terms of a sequence given its general term formula?

What are the differences between finite and infinite sequences?

What are the differences between finite and infinite sequences?

How can you write the explicit formula for a sequence that increases by a constant difference?

How can you write the explicit formula for a sequence that increases by a constant difference?

How do you represent sequences with alternating signs using explicit formulas?

How do you represent sequences with alternating signs using explicit formulas?

What is the general form of an exponential sequence and how do you find its terms?

What is the general form of an exponential sequence and how do you find its terms?

Your Beginning & Intermediate Algebra tutors

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

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

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