Question

Find the first six terms of each of the following sequences, starting with \(n=1\). $$ a_{n}=n^{2}-1 \text { for } n \geq 1 $$

Short answer

The first six terms are 0, 3, 8, 15, 24, and 35.

Step-by-step solution

Understand the Given Formula

The sequence is defined by the formula \( a_{n} = n^{2} - 1 \) for \( n \geq 1 \). This means each term in the sequence is found by substituting \( n \) with the term position.

Calculate the First Term (n=1)

Substitute \( n=1 \) into the formula: \( a_{1} = 1^2 - 1 = 0 \). The first term is 0.

Calculate the Second Term (n=2)

Substitute \( n=2 \) into the formula: \( a_{2} = 2^2 - 1 = 3 \). The second term is 3.

Calculate the Third Term (n=3)

Substitute \( n=3 \) into the formula: \( a_{3} = 3^2 - 1 = 8 \). The third term is 8.

Calculate the Fourth Term (n=4)

Substitute \( n=4 \) into the formula: \( a_{4} = 4^2 - 1 = 15 \). The fourth term is 15.

Calculate the Fifth Term (n=5)

Substitute \( n=5 \) into the formula: \( a_{5} = 5^2 - 1 = 24 \). The fifth term is 24.

Calculate the Sixth Term (n=6)

Substitute \( n=6 \) into the formula: \( a_{6} = 6^2 - 1 = 35 \). The sixth term is 35.

Compile the First Six Terms

The first six terms of the sequence are 0, 3, 8, 15, 24, and 35.

Key concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant difference to the previous term. However, it's essential to clarify that the given sequence in the example, based on the formula \( a_n = n^2 - 1 \), is not an arithmetic progression.
To be an arithmetic sequence, the difference between consecutive terms must always be the same. For instance, in a sequence such as 2, 5, 8, 11, each term increases by 3, making the common difference 3. In our provided sequence, 0, 3, 8, 15, 24, 35, the differences (3, 5, 7, 9, 11) are increasing. Due to this, we can determine that the sequence does not have a constant common difference and therefore is not an arithmetic progression.
Understanding the distinction between different types of sequences is crucial in tackling mathematical problems effectively. Always check whether the sequence has the same difference or if another type of progression is in play.

Sequence Formula

In mathematics, a sequence formula represents a rule defined to generate the terms of a sequence. These formulas are vital because they allow us to find any term in a sequence without listing all the terms manually.
In the original exercise, the formula provided is \( a_{n} = n^2 - 1 \). This formula is specific and depicts a sequence where each term is created by taking the square of \( n \) (the term's position in the sequence) and subtracting 1.
Understanding sequence formulas has significant advantages:

• Efficiency: Quickly calculate any term without computing all the preceding terms.
• Pattern Recognition: Identify the nature of the sequence, whether it's arithmetic, geometric, or neither.
• Problem Solving: Use it in broader mathematical contexts, including calculus and discrete mathematics.

Always begin by identifying what kind of sequence a particular formula represents. This helps approach problems with confidence and accuracy.

Term Calculation

Term calculation involves substituting a sequence term's position into the given formula to find the value of that term. It's a step-by-step procedure that provides beyond basics clarity regarding how each term is derived.
In our exercise, with the sequence formula \( a_{n} = n^2 - 1 \), calculating the terms is straightforward:

• For \( n=1 \): Substitute 1 into the formula, \( a_1 = 1^2 - 1 = 0 \). So, the first term is 0.
• For \( n=2 \): Substitute 2, \( a_2 = 2^2 - 1 = 3 \). The second term is 3.
• For \( n=3 \): Substitute 3, \( a_3 = 3^2 - 1 = 8 \). The third term is 8.
• For \( n=4 \): Substitute 4, \( a_4 = 4^2 - 1 = 15 \). The fourth term is 15.
• For \( n=5 \): Substitute 5, \( a_5 = 5^2 - 1 = 24 \). The fifth term is 24.
• For \( n=6 \): Substitute 6, \( a_6 = 6^2 - 1 = 35 \). The sixth term is 35.

Calculating terms from a sequence formula gives you each term's value quickly and ensures you correctly understand how sequences build over term positions. This builds a solid foundation for tackling more complex problems in algebra and beyond.