Problem 7 - A Graphical Approach to Precalculus with Limits

Question

Write the first five terms of each sequence. Do not use a calculator. $$a_{n}=\left(\frac{1}{3}\right)^{n}(n-1)$$

Step-by-Step Solution

Answer

The terms are 0, \( \frac{1}{9} \), \( \frac{2}{27} \), \( \frac{1}{27} \), and \( \frac{4}{243} \).

Step 1: Understand the Sequence Formula

The sequence is defined by the formula \( a_n = \left(\frac{1}{3}\right)^n (n-1) \). This means for any term \( a_n \), we multiply the \( n-1 \) by \( \left(\frac{1}{3}\right)^n \).

Step 2: Calculate the First Term

For \( n=1 \), substitute into the formula: \( a_1 = \left(\frac{1}{3}\right)^1 (1-1) = \left(\frac{1}{3}\right)^1 \times 0 = 0 \). Hence, the first term is 0.

Step 3: Calculate the Second Term

For \( n=2 \), substitute into the formula: \( a_2 = \left(\frac{1}{3}\right)^2 (2-1) = \left(\frac{1}{3}\right)^2 \times 1 = \frac{1}{9} \). Thus, the second term is \( \frac{1}{9} \).

Step 4: Calculate the Third Term

For \( n=3 \), substitute into the formula: \( a_3 = \left(\frac{1}{3}\right)^3 (3-1) = \left(\frac{1}{3}\right)^3 \times 2 = \frac{2}{27} \). Therefore, the third term is \( \frac{2}{27} \).

Step 5: Calculate the Fourth Term

For \( n=4 \), substitute into the formula: \( a_4 = \left(\frac{1}{3}\right)^4 (4-1) = \left(\frac{1}{3}\right)^4 \times 3 = \frac{3}{81} = \frac{1}{27} \). So, the fourth term is \( \frac{1}{27} \).

Step 6: Calculate the Fifth Term

For \( n=5 \), substitute into the formula: \( a_5 = \left(\frac{1}{3}\right)^5 (5-1) = \left(\frac{1}{3}\right)^5 \times 4 = \frac{4}{243} \). Hence, the fifth term is \( \frac{4}{243} \).

Key Concepts

Term Calculation Techniques

Understanding how to calculate terms in a sequence is crucial for identifying patterns and solving problems related to sequences. A sequence is generated by a specific formula, like the one given: \[ a_n = \left(\frac{1}{3}\right)^n (n-1) \]Here's a simplified way to think about calculating terms:

• Start by identifying what each part of the formula means. In the example formula, \( \left(\frac{1}{3}\right)^n \) indicates an exponential component, while \( (n-1) \) is a linear multiplier.
• Substitute values sequentially for \( n \) to find the corresponding terms.
• Always follow the order of operations - parentheses first, then exponents, followed by multiplication and division.

Besides just calculating the numbers, understanding how different terms in the formula interact can reveal patterns that might simplify the process or help you in problem-solving scenarios.
It's also useful to write out calculations step by step to avoid errors, especially with fractional and exponential components.

Understanding Exponential Sequences

Exponential sequences can seem daunting at first, but they can be simplified with understanding and practice. The sequence formula given\[ a_n = \left(\frac{1}{3}\right)^n (n-1) \]represents an exponential sequence.
The term \( \left(\frac{1}{3}\right)^n \) affects how quickly the values of the sequence grow or shrink. This exponential component:

• Controls the rate at which terms change - small changes in \( n \) can significantly impact the sequence's terms.
• Represents a geometric progression because each term after the first is the previous term multiplied by the base \( \frac{1}{3} \).
• Shows that as \( n \) gets larger, the terms' values reduce exponentially due to the base being a fraction less than 1.

Understanding these exponential trends will help you anticipate how terms behave in longer sequences. Additionally, recognizing these patterns offers a shortcut, allowing you to predict the nature of terms without heavy computation.

Step-by-Step Solutions Breakdown

When working through problems involving sequences, a step-by-step solution is invaluable. This method helps ensure accuracy and provides clarity at every stage. In solving the sequence:\[ a_n = \left(\frac{1}{3}\right)^n (n-1) \]Here’s how step-by-step aids in solutions:

• Step 1: Understand the formula. Knowing \( a_n = \left(\frac{1}{3}\right)^n (n-1) \) means recognizing how each part of the formula contributes to the term.
• Steps 2 to 6: These involve substituting values for \( n \) and computing using basic arithmetic operations. Each step focuses on calculating successive terms by plugging in \( n \) values sequentially.
• Breaking it down makes the problem simpler and prevents mixing of the sequence's parts.

Using a systematic approach not only makes complex problems more tractable but also builds confidence in handling sequences. Try writing out each step to see how calculations transform linearly or exponentially within sequences.