Problem 3 - Thomas Calculus

Question

Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4} .\) $$ a_{n}=\frac{(-1)^{n+1}}{2 n-1} $$

Step-by-Step Solution

Answer

The terms are \(a_1=1\), \(a_2=\frac{-1}{3}\), \(a_3=\frac{1}{5}\), \(a_4=\frac{-1}{7}\).

Step 1: Find the First Term

Substitute \( n = 1 \) into the formula \( a_n = \frac{(-1)^{n+1}}{2n - 1} \).\[ a_1 = \frac{(-1)^{1+1}}{2 \times 1 - 1} = \frac{1}{1} = 1 \]

Step 2: Find the Second Term

Substitute \( n = 2 \) into the formula \( a_n = \frac{(-1)^{n+1}}{2n - 1} \).\[ a_2 = \frac{(-1)^{2+1}}{2 \times 2 - 1} = \frac{-1}{3} \]

Step 3: Find the Third Term

Substitute \( n = 3 \) into the formula \( a_n = \frac{(-1)^{n+1}}{2n - 1} \).\[ a_3 = \frac{(-1)^{3+1}}{2 \times 3 - 1} = \frac{1}{5} \]

Step 4: Find the Fourth Term

Substitute \( n = 4 \) into the formula \( a_n = \frac{(-1)^{n+1}}{2n - 1} \).\[ a_4 = \frac{(-1)^{4+1}}{2 \times 4 - 1} = \frac{-1}{7} \]

Key Concepts

nth term calculation

When working with sequences in mathematics, an important concept is determining the nth term. The nth term of a sequence provides a formula that allows you to find any term in the sequence without listing all previous terms. In this exercise, the nth term is defined as:\[ a_n = \frac{(-1)^{n+1}}{2n - 1} \] This formula helps us calculate the sequence's elements at any position \( n \). Here’s how we calculate specific terms:

• Determine \( a_1 \) by plugging \( n = 1 \) into the formula. Simplifying gives us \( a_1 = 1 \).
• Determine \( a_2 \) by plugging \( n = 2 \) into the formula. Simplifying gives us \( a_2 = -\frac{1}{3} \).
• Continue this process for \( a_3 \) and \( a_4 \), obtaining \( a_3 = \frac{1}{5} \) and \( a_4 = -\frac{1}{7} \).

By using the nth term formula, you can find any term in the sequence quickly and accurately.

alternating sequence

Alternating sequences have a pattern in which the sign of the terms changes regularly, typically between positive and negative. This regular alternation is key in sequences like the one in the exercise. The sequence \( \left\{a_n\right\} \) is an example of an alternating sequence, highlighted by the term \((-1)^{n+1}\).
Here's how the alternating pattern works:

• The expression \((-1)^{n+1}\) determines the sign of each term.
• When \( n \) is even, \((-1)^{n+1}\) becomes negative, leading the term to be negative.
• Conversely, when \( n \) is odd, \((-1)^{n+1}\) becomes positive, leading the term to be positive.

This pattern creates a back-and-forth shift in the signs, forming the essence of an alternating sequence and offering insight into the sequence's overall behavior.

substitution method

The substitution method is a straightforward mathematical technique used to find specific values in sequences. It involves substituting a particular value, typically an integer, for a variable in a given formula. This allows for direct calculation of the sequence terms.
Here’s how it applies to our exercise:

• To find \( a_1 \), substitute \( n = 1 \). You replace \( n \) with 1 in the formula and simplify the arithmetic, resulting in \( a_1 = 1 \).
• To find \( a_2 \), substitute \( n = 2 \). Follow the same process, giving \( a_2 = -\frac{1}{3} \).
• This method is applied similarly for \( a_3 \) and \( a_4 \), making it easy to calculate each term directly from the general formula.

By mastering the substitution method, you can efficiently find individual terms in any sequence without needing to compute them sequentially from the start.