Sequences: Videos & Practice Problems

Memorizing numbers, such as bank account numbers or phone numbers, often requires understanding their order. This concept is foundational to sequences, which are essentially lists of numbers arranged in a specific order. For instance, the sequence 2, 4, 6, 8 illustrates how each number is a term, also known as an element or member of the sequence. Recognizing patterns within sequences is crucial; in the example of 2, 4, 6, 8, each term increases by 2. To find the next term, simply add 2 to the last term, resulting in 10.

Sequences can be classified as finite or infinite. A finite sequence has a specific endpoint, while an infinite sequence continues indefinitely. For example, the sequence 3, 6, 9, 12, 18 is finite because it ends at 18, whereas a sequence represented as 1, 1/2, 1, ... continues without end, indicated by the ellipsis.

Sequences share similarities with functions, as both follow specific rules and can be expressed with equations. However, the key difference lies in their inputs. In functions, inputs can be any real number, while sequences use integer indexes, denoted by the letter n, starting from 1 and increasing by 1. For example, in the equation \( a_n = 2n \), the outputs (terms of the sequence) are calculated by substituting integer values for n, yielding the sequence 2, 4, 6, 8, 10.

To illustrate further, consider the formula \( a_n = n^2 \). By substituting n with 1, 2, and 3, we find the first three terms of the sequence: \( a_1 = 1^2 = 1 \), \( a_2 = 2^2 = 4 \), and \( a_3 = 3^2 = 9 \). Similarly, for the formula \( a_n = \frac{1}{n} + 3 \), substituting n gives \( a_1 = \frac{1}{1} + 3 = 4 \), \( a_2 = \frac{1}{2} + 3 = 3.5 \), and \( a_3 = \frac{1}{3} + 3 \approx 3.33 \).

Sequences can also involve negative numbers and exponents. For example, in the equation \( a_n = (-1)^n \), the terms alternate between -1 and 1, depending on whether n is odd or even. Thus, \( a_1 = (-1)^1 = -1 \), \( a_2 = (-1)^2 = 1 \), and \( a_3 = (-1)^3 = -1 \).

Understanding sequences is essential for recognizing patterns and solving problems in mathematics. With practice, identifying terms and their relationships becomes more intuitive, paving the way for deeper mathematical exploration.

Understanding sequences involves recognizing patterns and deriving explicit formulas that represent these patterns mathematically. An explicit formula is a function of the term index \( n \) that allows you to calculate the term directly. The key to finding these formulas lies in identifying the relationships between the terms in the sequence.

One common type of sequence 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 of the sequence. In this case, since the first term is 5, we find that \( c = 4 \), leading to the formula:

\[ a_n = n + 4 \]

Another type of sequence is one that alternates in sign, such as \( -5, 5, -5, 5 \). This can be represented using the expression \( (-1)^n \), which alternates between -1 and 1 based on whether \( n \) is odd or even. To adjust for the specific values in the sequence, we multiply by 5:

\[ a_n = (-1)^n \cdot 5 \]

Sequences can also involve fractions, such as \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \). Here, the numerators follow the pattern \( n \), while the denominators increase by 1 starting from 2. Thus, the general formula becomes:

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

Exponential sequences, like \( 2, 4, 8, 16, 32 \), grow at an increasing rate. The general formula for an exponential sequence can be expressed as:

\[ a_n = b^n \]

In this case, since the base is 2, the formula is:

\[ a_n = 2^n \]

By recognizing these patterns and applying the appropriate adjustments, you can derive the explicit formulas for various types of sequences. This skill is essential for solving problems related to sequences and series in mathematics.

In this example, we explore how to derive a general formula for a sequence based on its given terms. The sequence provided consists of fractions: 1/(1×2), 1/(2×3), 1/(3×4), 1/(4×5). To find the general term, denoted as a_n, we analyze the patterns in the numerators and denominators of these fractions.

First, we observe that the numerator of each term is consistently 1, indicating that it does not depend on the term index n. Therefore, the numerator in our general formula will simply be 1.

Next, we examine the denominator, which consists of products of two consecutive integers: 1×2, 2×3, 3×4, 4×5. The first part of the denominator corresponds to the term index n, while the second part is always one greater than n. This leads us to conclude that the denominator can be expressed as n(n + 1).

Thus, the general formula for the sequence can be written as:

a_n = 1/(n(n + 1))

To find a specific term in the sequence, we can substitute a value for n. For example, to find the 15th term, we substitute n = 15 into our formula:

a_{15} = 1/(15(15 + 1)) = 1/(15 × 16)

Calculating this gives:

a_{15} = 1/240, which is approximately 0.0041667.

This method allows us to quickly compute any term in the sequence without needing to list all preceding terms, demonstrating the power of identifying patterns and formulating general expressions in sequences.

In this discussion, we explore the process of deriving a general formula for a sequence of numbers: -2, 4, -6, 8, -10, and so forth. The goal is to create a formula that allows us to calculate the nth term efficiently, especially for larger values of n, such as 18.

First, we observe the pattern in the sequence. The numbers alternate in sign, which indicates that we can use the expression (-1)^n to account for this alternation. Specifically, when n is odd, the term will be negative, and when n is even, it will be positive. This is crucial for accurately representing the sequence.

Next, we note that the absolute values of the terms increase by 2 each time. To represent this linear growth, we can multiply n by 2, leading us to the expression 2n. Thus, the general formula for the nth term of the sequence can be combined as follows:

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

To verify this formula, we can calculate the first few terms:

• 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 our formula accurately represents the sequence. Now, to find the 18th term, we substitute n with 18 in the formula:

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

Since 18 is even, (-1)^{18} = 1, leading to:

a_{18} = 1 \cdot 36 = 36

Thus, the 18th term of the sequence is 36. This general formula not only simplifies the calculation of terms in the sequence but also highlights the importance of recognizing patterns in sequences to derive effective mathematical expressions.

Arithmetic sequences are a specific type of sequence where the difference between consecutive terms remains constant. This constant difference is referred to as the common difference, denoted by the letter d. For example, in the sequence 2, 7, 12, 17, each term increases by 5, making d = 5.

To express an arithmetic sequence using a recursive formula, we can define the next term based on the previous term. The general form of the recursive formula is:

\( a_n = a_{n-1} + d \)

In this case, to find the next term, you simply add the common difference to the previous term. For the sequence mentioned, the recursive formula would be:

\( a_n = a_{n-1} + 5 \)

To illustrate this, if we start with \( a_1 = 2 \), we can calculate the subsequent terms:

• \( a_2 = a_1 + 5 = 2 + 5 = 7 \)
• \( a_3 = a_2 + 5 = 7 + 5 = 12 \)
• \( a_4 = a_3 + 5 = 12 + 5 = 17 \)

In another example, if we have a sequence defined by a common difference of -6, starting with \( a_1 = 9 \), the recursive formula would be:

\( a_n = a_{n-1} - 6 \)

Calculating the terms gives us:

• \( a_2 = 9 - 6 = 3 \)
• \( a_3 = 3 - 6 = -3 \)
• \( a_4 = -3 - 6 = -9 \)

When tasked with writing a recursive formula from a given sequence, the first step is to determine the common difference by subtracting any two consecutive terms. For instance, in the sequence 2, 5, 8, 11, 14, the common difference is:

\( d = 5 - 2 = 3 \)

Thus, the recursive formula becomes:

\( a_n = a_{n-1} + 3 \)

To complete the recursive formula, it is essential to specify the initial term, which in this case is \( a_1 = 2 \). Therefore, the complete recursive formula is:

\( a_n = a_{n-1} + 3 \) with \( a_1 = 2 \)

Understanding how to derive and utilize recursive formulas for arithmetic sequences allows for the generation of any term in the sequence based on the initial term and the common difference.

In arithmetic sequences, understanding how to derive both recursive and general formulas is essential for efficiently calculating terms. A recursive formula defines each term based on the previous term, while a general formula allows for direct calculation of any term in the sequence without needing prior terms.

The general formula for the nth term of an arithmetic sequence can be expressed as:

$$ a_n = a_1 + d \cdot (n - 1) $$

Here, a_n represents the nth term, a_1 is the first term, d is the common difference between consecutive terms, and n is the term's position in the sequence.

To illustrate, consider the sequence 2, 7, 12, 17. The first term a_1 is 2, and the common difference d is 5 (since 7 - 2 = 5). Plugging these values into the general formula gives:

$$ a_n = 2 + 5 \cdot (n - 1) $$

Using this formula, we can find any term in the sequence. For example, to find the 4th term:

$$ a_4 = 2 + 5 \cdot (4 - 1) = 2 + 5 \cdot 3 = 2 + 15 = 17 $$

This confirms that the 4th term is indeed 17, as expected.

Now, let’s apply this to another sequence: 2, 5, 8, 11, 14. Here, a_1 is 2, and the common difference d is 3. The general formula becomes:

$$ a_n = 2 + 3 \cdot (n - 1) $$

To find the 101st term, substitute n with 101:

$$ a_{101} = 2 + 3 \cdot (101 - 1) = 2 + 3 \cdot 100 = 2 + 300 = 302 $$

This method demonstrates the power of general formulas in arithmetic sequences, allowing for quick calculations of terms without the need for extensive recursive calculations.

In mathematics, particularly in sequences, converting a recursive formula into a general formula can significantly simplify calculations. For an arithmetic sequence, where each term is derived from the previous term by adding a constant, this process is straightforward.

Consider a recursive formula where the next term is the previous term plus a common difference, denoted as d. For example, if the starting term is 2 and the common difference is 3, the recursive relationship can be expressed as:

Recursive Formula: a_n = a_n-1 + d, where a₁ = 2 and d = 3.

To derive the general formula for the nth term of the sequence, the formula can be expressed as:

General Formula: a_n = a₁ + d(n - 1).

In this case, substituting the known values gives:

a_n = 2 + 3(n - 1).

This formula allows for the calculation of any term in the sequence without needing to compute all previous terms. For instance, to find the 15th term, substitute n = 15 into the general formula:

a₁₅ = 2 + 3(15 - 1).

Calculating this yields:

a₁₅ = 2 + 3(14) = 2 + 42 = 44.

Thus, the 15th term of the sequence is 44. This method of deriving a general formula from a recursive one is a powerful tool in sequence analysis, allowing for quick calculations of any term in the sequence.

In this problem, we are tasked with finding the general term of a sequence given two specific terms: \( a_4 = -2 \) and \( a_6 = 6 \). To approach this, we utilize the general formula for an arithmetic sequence, which is expressed as:

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

Here, \( a_1 \) represents the first term of the sequence, and \( d \) is the common difference between consecutive terms. Since we are given non-consecutive terms, we need to derive a system of equations to find both \( a_1 \) and \( d \).

Substituting \( n = 4 \) into the general formula gives us:

\( a_4 = a_1 + d(4 - 1) \Rightarrow -2 = a_1 + 3d \quad (1) \

For \( n = 6 \), we have:

\( a_6 = a_1 + d(6 - 1) \Rightarrow 6 = a_1 + 5d \quad (2) \

Now we have a system of two equations:

1. \( a_1 + 3d = -2 \)

2. \( a_1 + 5d = 6 \)

To solve this system, we can use the elimination method. By subtracting equation (1) from equation (2), we eliminate \( a_1 \):

\( (a_1 + 5d) - (a_1 + 3d) = 6 - (-2) \)

This simplifies to:

\( 2d = 8 \)

Dividing both sides by 2 gives us:

\( d = 4 \)

Now that we have the common difference \( d \), we can substitute it back into either equation to find \( a_1 \). Using equation (1):

\( a_1 + 3(4) = -2 \)

This simplifies to:

\( a_1 + 12 = -2 \)

Subtracting 12 from both sides yields:

\( a_1 = -14 \)

With both \( a_1 \) and \( d \) determined, we can now write the general formula for the sequence:

\( a_n = -14 + 4(n - 1) \)

To find a specific term, such as \( a_{18} \), we substitute \( n = 18 \) into the general formula:

\( a_{18} = -14 + 4(18 - 1) \)

This simplifies to:

\( a_{18} = -14 + 4 \times 17 \)

Calculating \( 4 \times 17 \) gives 68, so:

\( a_{18} = -14 + 68 = 54 \)

Thus, the 18th term of the sequence is \( 54 \). This method illustrates how to derive a general formula from known terms in an arithmetic sequence by forming and solving a system of equations.

In mathematics, sequences can be categorized into different types based on how the terms relate to one another. One such type is the arithmetic sequence, where the difference between consecutive terms remains constant. For example, in the sequence 3, 6, 9, 12, the common difference is 3. In contrast, a geometric sequence features a constant ratio between consecutive terms, rather than a constant difference. An example of a geometric sequence is 3, 9, 27, 81, where each term is obtained by multiplying the previous term by a common ratio, denoted as r.

In the given geometric sequence, the common ratio r is equal to 3, as each term is three times the previous one. This can be expressed using a recursive formula, which defines each term based on the preceding term. The general structure of a recursive formula for a geometric sequence is:

\( a_n = a_{n-1} \times r \)

Here, \( a_n \) represents the new term, \( a_{n-1} \) is the previous term, and r is the common ratio. This formula highlights that to find the next term in a geometric sequence, you multiply the previous term by the common ratio.

For example, consider the sequence 5, 20, 80, 320. To determine the common ratio, you can divide any two consecutive terms. From 5 to 20, you multiply by 4, and this pattern continues for the subsequent terms. Thus, the common ratio r is 4. The recursive formula for this sequence can be written as:

\( a_n = a_{n-1} \times 4 \)

Additionally, it is essential to specify the first term of the sequence, which in this case is \( a_1 = 5 \). This recursive formula allows you to generate the entire sequence by repeatedly applying the multiplication operation.

In summary, while arithmetic sequences grow linearly through addition, geometric sequences exhibit exponential growth through multiplication, leading to much faster increases in their terms. Understanding these differences is crucial for effectively working with various types of sequences in mathematics.

In mathematics, understanding sequences is crucial, particularly when distinguishing between arithmetic and geometric sequences. A geometric sequence is defined by a common ratio, which is the factor by which each term is multiplied to obtain the next term. To find any term in a geometric sequence without relying on previous terms, we use a general formula.

The general formula for the nth term of a geometric sequence can be expressed as:

$$ a_n = a_1 \cdot r^{n-1} $$

Here, a_n represents the nth term, a_1 is the first term of the sequence, r is the common ratio, and n is the term number. This formula allows for direct calculation of any term in the sequence.

For example, consider the geometric sequence 3, 6, 12, 24. The first term a_1 is 3, and the common ratio r is 2 (since each term is multiplied by 2 to get the next). Using the general formula, we can express the nth term as:

$$ a_n = 3 \cdot 2^{n-1} $$

To find the fourth term (n = 4), we substitute into the formula:

$$ a_4 = 3 \cdot 2^{4-1} = 3 \cdot 2^3 = 3 \cdot 8 = 24 $$

This confirms that the fourth term is indeed 24, as expected.

Now, let’s apply this to another example with the sequence 5, 20, 80, 320. Here, the first term a_1 is 5, and the common ratio r is 4. The general formula for this sequence is:

$$ a_n = 5 \cdot 4^{n-1} $$

To find the twelfth term (n = 12), we substitute:

$$ a_{12} = 5 \cdot 4^{12-1} = 5 \cdot 4^{11} $$

Calculating this gives a very large number, specifically 20,917,520. This illustrates the power of the general formula in efficiently determining terms in a geometric sequence, especially for higher indices where recursive methods would be cumbersome.

In summary, the general formula for a geometric sequence is a powerful tool that simplifies the process of finding terms, allowing for quick calculations based on the first term and the common ratio.