Sequences: Videos & Practice Problems

Understanding sequences is essential in mathematics, as they represent ordered lists of numbers, known as terms. Each term in a sequence is identified by its position, such as the first term, second term, and so on. For example, the sequence 2, 4, 6, 8 illustrates a clear pattern where each term increases by 2. To find the fifth term, one simply adds 2 to the last known 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 instance, the sequence 3, 6, 9, 12, 15, 18 is finite, as it ends at 18. In contrast, the sequence 1, 1/2, 1/3, 1/4, ... is infinite, indicated by the ellipsis suggesting it continues forever.

Sequences share similarities with functions, as both follow specific rules and can be expressed mathematically. However, the key difference lies in their inputs. In functions, inputs can be any real number, while in sequences, inputs are restricted to positive integers, known as indexes, represented by the letter n. For example, in the sequence defined by the formula \( a_n = 2n \), the first five terms can be calculated by substituting n with 1, 2, 3, 4, and 5, yielding the terms 2, 4, 6, 8, and 10.

To further illustrate, consider the formula \( a_n = n^2 \). The first three terms can be calculated as follows: for \( n = 1 \), \( a_1 = 1^2 = 1 \); for \( n = 2 \), \( a_2 = 2^2 = 4 \); and for \( n = 3 \), \( a_3 = 3^2 = 9 \). Thus, the first three terms of this sequence are 1, 4, and 9.

Another example is the sequence defined by \( a_n = \frac{1}{n + 3} \). The first term is \( a_1 = \frac{1}{1 + 3} = \frac{1}{4} \), the second term is \( a_2 = \frac{1}{2 + 3} = \frac{1}{5} \), and the third term is \( a_3 = \frac{1}{3 + 3} = \frac{1}{6} \). The first three terms are therefore \( \frac{1}{4}, \frac{1}{5}, \frac{1}{6} \).

Lastly, consider the sequence defined by \( a_n = (-1)^n \). The first term is \( a_1 = (-1)^1 = -1 \), the second term is \( a_2 = (-1)^2 = 1 \), and the third term is \( a_3 = (-1)^3 = -1 \). This sequence alternates between -1 and 1, demonstrating that sequences can also include negative numbers.

In summary, sequences are fundamental mathematical constructs that consist of ordered terms, which can be finite or infinite. They can be expressed through formulas, allowing for the calculation of specific terms based on their position in the sequence. Understanding these concepts is crucial for further studies in mathematics.

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.

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.

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.

In mathematics, sequences can be defined using different types of formulas, primarily general formulas and recursive formulas. A general formula provides a direct way to calculate the nth term of a sequence based on its index, n. For example, if we have a sequence defined by the formula \( a_n = 2n \), we can easily find the first five terms by substituting n with 1, 2, 3, 4, and 5, yielding the terms 2, 4, 6, 8, and 10.

On the other hand, a recursive formula defines each term based on the previous term(s) in the sequence. For instance, a recursive formula might be expressed as \( a_n = a_{n-1} + 2 \). Here, \( a_{n-1} \) represents the previous term, and to find the next term, you simply add 2 to it. Starting with the first term \( a_1 = 2 \), you can calculate subsequent terms: \( a_2 = a_1 + 2 = 4 \), \( a_3 = a_2 + 2 = 6 \), and so forth, resulting in the same sequence of 2, 4, 6, 8, and 10.

The key distinction between general and recursive formulas lies in their approach: general formulas require knowledge of n to compute the nth term, while recursive formulas rely on the previous term to derive the next one. This makes recursive formulas particularly useful when only a few terms need to be calculated, as they can simplify the process without needing to derive a general formula.

For example, consider a recursive formula defined as \( a_n = 2 \cdot a_{n-1} + 3 \) with \( a_1 = 1 \). To find the second term, you would calculate \( a_2 = 2 \cdot a_1 + 3 = 2 \cdot 1 + 3 = 5 \). Continuing this process, the third term would be \( a_3 = 2 \cdot a_2 + 3 = 2 \cdot 5 + 3 = 13 \), and the fourth term would be \( a_4 = 2 \cdot a_3 + 3 = 2 \cdot 13 + 3 = 29 \).

Another example could involve a recursive formula like \( a_n = n \cdot a_{n-1} \) with \( a_1 = 1 \). To find the second term, you would compute \( a_2 = 2 \cdot a_1 = 2 \cdot 1 = 2 \). For the third term, \( a_3 = 3 \cdot a_2 = 3 \cdot 2 = 6 \), and for the fourth term, \( a_4 = 4 \cdot a_3 = 4 \cdot 6 = 24 \).

In summary, both general and recursive formulas are valuable tools for defining sequences, each serving different purposes depending on the context and requirements of the problem at hand. Understanding how to utilize both types of formulas enhances your ability to work with sequences effectively.