Problem 22 - Thomas Calculus

Question

Find a formula for the \(n\)th term of the sequence. $$ 2,6,10,14,18, \dots $$

Step-by-Step Solution

Answer

The formula for the nth term is \( a_n = 4n - 2 \).

Step 1: Identify the Pattern

Look at the sequence and determine how each term is related to the previous term. The given terms are 2, 6, 10, 14, 18.The difference between consecutive terms is:\[ 6 - 2 = 4, \]\[ 10 - 6 = 4, \]\[ 14 - 10 = 4, \]\[ 18 - 14 = 4. \] Thus, each term increases by 4.

Step 2: Construct the General Formula for an Arithmetic Sequence

Since the sequence has a common difference of 4, it is an arithmetic sequence.The general formula for an arithmetic sequence is:\[ a_n = a_1 + (n-1) imes d, \]where \( a_n \) is the \( n \)th term, \( a_1 \) is the first term, and \( d \) is the common difference.

Step 3: Substitute the Known Values into the Formula

Substitute the first term and the common difference into the formula:\[ a_1 = 2, \]\[ d = 4. \]The formula becomes:\[ a_n = 2 + (n-1) imes 4. \]

Step 4: Simplify the Formula

Simplify the expression to find a cleaner formula for the \( n \)th term:\[ a_n = 2 + 4(n-1) = 2 + 4n - 4. \]Simplifying further gives:\[ a_n = 4n - 2. \]

Step 5: Verify the Formula

Use the formula to find the first few terms and ensure they match the original sequence:For \( n = 1 \), \( a_1 = 4(1) - 2 = 2 \).For \( n = 2 \), \( a_2 = 4(2) - 2 = 6 \).For \( n = 3 \), \( a_3 = 4(3) - 2 = 10 \).For \( n = 4 \), \( a_4 = 4(4) - 2 = 14 \).For \( n = 5 \), \( a_5 = 4(5) - 2 = 18 \).The formula correctly produces the sequence terms.

Key Concepts

Common Difference

The common difference in an arithmetic sequence is the constant amount by which successive terms increase or decrease. In our given sequence, we can see the numbers 2, 6, 10, 14, and 18. By observing the pattern:

• Subtract 2 from 6, yielding 4.
• Subtract 6 from 10, resulting in 4.
• Subtract 10 from 14, again resulting in 4.
• Finally, subtract 14 from 18, which is 4 once more.

These calculations illustrate that the common difference, denoted by \(d\), is 4.
Recognizing this difference helps in constructing formulas for sequences.
Knowing the common difference is crucial as it helps to predict the subsequent terms in the sequence.

nth Term Formula

The formula for the \(n\)th term of an arithmetic sequence allows us to find any term in a sequence without listing all prior terms. The generalized form is:\[ a_n = a_1 + (n-1) \times d \]where:

• \(a_n\) is the \(n\)th term you are trying to find.
• \(a_1\) is the first term of the sequence.
• \(d\) is the common difference.

In our specific problem:

• \(a_1 = 2\)
• \(d = 4\)

By inserting these values into the formula, we get the expression: \[ a_n = 2 + (n-1) \times 4 \]This equation helps you efficiently compute any term given its position \(n\). Learning how to use the \(n\)th term formula expands your ability to solve real-world problems involving sequences.

Sequence Pattern

Recognizing patterns in a sequence is key to understanding how it behaves. In arithmetic sequences, the pattern involves a constant addition that is predictable.
In our example sequence of 2, 6, 10, 14, 18:

• The pattern is an increase by 4 with each subsequent term.
• This is consistent across the sequence and can be represented by the common difference.

Identifying the sequence pattern is essential because:

• It aids in quickly diagnosing the rule governing the sequence.
• Helps in constructing a general formula like the \(n\)th term formula.

Knowing the sequence pattern not only saves time but boosts confidence in solving similar problems.

Thomas Calculus

Thomas Calculus is a renowned textbook that often introduces foundational mathematical concepts, such as sequences and series, to help students develop a deeper understanding of calculus.
While it primarily focuses on calculus, the book often delves into sequences to lay the groundwork for more advanced topics like series.

• Thomas Calculus teaches sequences as a lead-in to understanding limits, continuity, and other core calculus concepts.
• Examining arithmetic sequences in this context shows how seemingly simple patterns form the backbone of more complex mathematical ideas.

By comprehending sequences with Thomas Calculus, you build a robust mathematical foundation.
This makes tackling calculus concepts more approachable and intuitive.