Problem 83
Question
Growth of an Exponential Function Suppose you are offered a job that lasts one month, and you are to be very well paid. Which of the following methods of payment is more profitable for you? (a) One million dollars at the end of the month (b) Two cents on the first day of the month, 4 cents on the second day, 8 cents on the third day, and, in general, \(2^{n}\) cents on the \(n\) th day
Step-by-Step Solution
Verified Answer
Option (b) is more profitable, yielding $21,474,836.46.
1Step 1: Understanding the Problem
We need to determine which payment option is more profitable over the course of 30 days: receiving one million dollars at the end of the month or receiving an exponentially increasing amount starting with 2 cents on the first day and doubling each subsequent day.
2Step 2: Calculate Sum for Option (b)
The payment on the nth day is given by the expression \(2^n\) cents. We need to find the total amount after 30 days. The formula for the sum of a geometric series is \(S = a \frac{r^n - 1}{r - 1}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. Here, \(a = 2\), \(r = 2\), and \(n = 30\). We find the sum in cents: \(S = 2 \frac{2^{30} - 1}{2 - 1}\).
3Step 3: Calculate Sum Value
Now compute \(S = 2 \times (2^{30} - 1)\). Using a calculator, \(2^{30} = 1,073,741,824\). Thus, the sum is \(2 \times (1,073,741,824 - 1) = 2,147,483,646\) cents, which equals \$21,474,836.46.
4Step 4: Comparison of Options
Compare \\(21,474,836.46 from option (b) to \\)1,000,000 from option (a). Clearly, option (b) yields a higher amount by the end of the month.
5Step 5: Conclusion
Option (b), the doubling cents each day, results in \\(21,474,836.46, which is much greater than the \\)1,000,000 offered in option (a). Thus, option (b) is more profitable.
Key Concepts
Geometric SeriesSum of a Geometric SeriesProblem-Solving in Mathematics
Geometric Series
A geometric series is a way to add up numbers, where each term after the first is a constant multiple of the previous term. For example, if you start with 2 and multiply each subsequent term by 2, you get a series like 2, 4, 8, 16, and so on. Such sequences are quite common and are extremely useful in various branches of mathematics and problem-solving.
The general form of a geometric series can be described as follows:
Understanding how geometric series works is essential when dealing with exponentially growing problems, like choosing between a lump sum and a doubling payment plan.
The general form of a geometric series can be described as follows:
- First term: \(a\)
- Common ratio: \(r\)
- Number of terms: \(n\)
Understanding how geometric series works is essential when dealing with exponentially growing problems, like choosing between a lump sum and a doubling payment plan.
Sum of a Geometric Series
To find the total of a geometric series, you'd use what's called the "sum formula." This formula allows you to quickly compute the sum of all the terms in a series without adding each term individually.
The formula for the sum of the first \(n\) terms of a geometric series is\[ S = a \frac{r^n - 1}{r - 1} \]where \(S\) is the sum, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.
In our exercise, option (b) involves doubling the amount over 30 days. We applied this sum formula with \(a = 2\), \(r = 2\), and \(n = 30\), resulting in a vast sum far exceeding the one million dollar option. Calculating these sums can give you insights into how quickly numbers can grow with exponential increases.
The formula for the sum of the first \(n\) terms of a geometric series is\[ S = a \frac{r^n - 1}{r - 1} \]where \(S\) is the sum, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.
In our exercise, option (b) involves doubling the amount over 30 days. We applied this sum formula with \(a = 2\), \(r = 2\), and \(n = 30\), resulting in a vast sum far exceeding the one million dollar option. Calculating these sums can give you insights into how quickly numbers can grow with exponential increases.
Problem-Solving in Mathematics
Problem-solving is a fundamental skill in mathematics that involves understanding the problem, setting up equations, and finding solutions. It often includes breaking down complex problems into simpler steps or using formulas like the geometric series sum for efficient solutions.
In the given exercise, we faced a problem of choosing the more profitable payment method. The problem was solved by first understanding and calculating the sum of the geometric series accurately. This allowed us to compare mathematical outcomes effectively.
In the given exercise, we faced a problem of choosing the more profitable payment method. The problem was solved by first understanding and calculating the sum of the geometric series accurately. This allowed us to compare mathematical outcomes effectively.
- Understanding the problem: Analyzing what is given and what needs to be found.
- Choosing formulas and methods: Selecting the appropriate mathematical tools—here, the sum of a geometric series.
- Performing calculations: Executing the necessary calculations to arrive at a solution.
- Comparing results: Evaluating which option offers greater value.
Other exercises in this chapter
Problem 83
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