Problem 70 - A Graphical Approach to Precalculus with Limits

Question

Determine the largest value of \(n\) that satisfies the inequality. $$\sum_{k=1}^{n} 3(0.5)^{k} \leq 2.8$$

Step-by-Step Solution

Answer

The largest value of n is 3.

Step 1: Understanding the Summation

The expression \( \sum_{k=1}^{n} 3(0.5)^k \) is a geometric series with the first term \( a = 3(0.5)^1 = 1.5 \) and the common ratio \( r = 0.5 \). Our task is to find the largest integer \( n \) such that the sum of this series is less than or equal to 2.8.

Step 2: Formula for the Sum of a Finite Geometric Series

The sum of the first \( n \) terms of a geometric series is given by \( S_n = a \frac{1-r^n}{1-r} \). In this case, \( a = 1.5 \) and \( r = 0.5 \), hence, \[ S_n = 1.5 \frac{1-(0.5)^n}{0.5} = 3(1-(0.5)^n). \]

Step 3: Set Up the Inequality

We need to solve the inequality \[ 3(1-(0.5)^n) \leq 2.8. \] First, simplify the inequality by dividing both sides by 3:\[ 1 - (0.5)^n \leq \frac{2.8}{3}. \]

Step 4: Solve the Simplified Inequality

Next, calculate \( \frac{2.8}{3} = 0.9333\ldots \) which can be approximately considered as 0.933. Subtract 1 from both sides of the inequality \[ -(0.5)^n \leq -0.067 \] and multiplying by -1, we get\[ (0.5)^n \geq 0.067. \]

Step 5: Finding the Largest n

To find the largest \( n \), use logarithms. Taking the logarithm (base 10) of both sides gives:\[ n \log_{10}(0.5) \geq \log_{10}(0.067). \] This simplifies to \[ n \leq \frac{\log_{10}(0.067)}{\log_{10}(0.5)}. \] Calculate these values: \( \log_{10}(0.067) \approx -1.173 \) and \( \log_{10}(0.5) \approx -0.301 \). Thus, \( n \leq \frac{-1.173}{-0.301} \approx 3.90. \) Since \( n \) must be an integer, the largest possible value is \( n = 3. \)

Key Concepts

Geometric Series

A geometric series is a sequence of terms where each term after the first is found by multiplying the previous term by a fixed number known as the common ratio. The series can be represented as:

• First term: \( a \)
• Common ratio: \( r \)
• General term: \( ar^{n-1} \)

For our series, the first term is \( a = 3(0.5)^1 = 1.5 \), and the common ratio \( r = 0.5 \). This pattern continues for each successive term in the series. Understanding this setup helps in calculating the sum of a finite geometric series, which in turn aids in solving inequalities involving series sums.

Inequalities

Inequalities represent a range of values rather than a specific number, showing that one side is either less than, greater than, equal to, or a combination with another value. In this task, the inequality is expressed as the sum of a geometric series.

• Given inequality: \( 3(1-(0.5)^n) \leq 2.8 \)
• This equation sets an upper limit for the range of sum values \( S_n \).
• Main goal: find the largest integer \( n \) that satisfies this condition.

Simplifying inequalities often involves dividing through by constants, moving terms, and manipulating the inequality sign, as seen by dividing by 3 and reversing the inequality when multiplying by negative values.

Logarithms

Logarithms are the opposite operation of exponentials and are useful for solving equations where the variable is in an exponent. In simpler terms, a logarithm answers the question: "To what power must we raise one number to obtain another number?"

• Notation: \( \, \log_b(x) \)
• In this exercise, we use base 10 log for convenience.

To solve the inequality \( (0.5)^n \geq 0.067 \), logarithms are utilized to isolate \( n \). After taking logs, the inequality becomes easier to manipulate, leading us to find \( n \leq \frac{\log_{10}(0.067)}{\log_{10}(0.5)} \). This technique provides the solution to finding the largest possible integer value of \( n \).

Summation

Summation is a way of adding up a sequence of numbers, typically expressed with the Greek letter sigma \( \Sigma \). It provides a compact way to express large sums. For a geometric series, the sum of the first \( n \) terms is given by the formula:

• \( S_n = a \frac{1-r^n}{1-r} \)
• Where \( a \) is the first term and \( r \) is the common ratio.

In the exercise: \( S_n = 3(1-(0.5)^n) \). We used this formula to set up the inequality \( 3(1-(0.5)^n) \leq 2.8 \), leading us to find \( n \). Summation is a key tool for handling cumulative totals in series, making it vital for resolving combinations and limits efficiently.