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Magazine Chapter 10: Problem 58
PERSONAL FINANCE: Payments (Graphing calculator with series operations helpful) A person gives you \(\$ 1,\) waits 1 minute, gives you another dollar, waits 2 minutes, gives you another dollar, waits four minutes, etc., doubling the wait each time. How much money will you have at the end of the first year?
Short Answer
Expert verified
You will have $19 by the end of the first year.
Step by step solution
01
Understand the Problem
The person is following a sequence where they give you $1, wait 1 minute, give you another $1, and then double the wait time each time. Your task is to find out how much money you will have by the end of the first year (525,600 minutes).
02
Identify the Pattern
The waiting time follows a geometric sequence: 1, 2, 4, 8, etc., doubling each time. The general term for the waiting time is given by \(2^{n-1}\) where \(n\) is the nth payment.
03
Find the Number of Payments
Let's find the maximum \(n\) for which the total waiting time is less than or equal to 525,600 minutes. We need the sum of the waiting times \(S_n = 1 + 2 + 4 + \, ... \, + 2^{n-1}\) to be less than or equal to 525,600.
04
Use Geometric Series Sum Formula
The sum of the first \(n\) terms of a geometric sequence is \(S_n = 2^n - 1\). Solve \(2^n - 1 \leq 525,600\) to find the largest integer \(n\).
05
Solve for n
Solve \(2^n \leq 525,601\). Perform the calculation: \(2^{19} = 524,288\) and \(2^{20} = 1,048,576\). So, the largest \(n\) is 19 as \(2^{19} < 525,600 < 2^{20}\).
06
Determine Total Sum of Money Received
With \(n=19\), there will be 19 payments by the end of the year. Since each payment is $1, the total money received is \(19\) dollars.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In our exercise, this concept was key to understanding the sequence of waiting times: 1 minute, 2 minutes, 4 minutes, and so on. Here, the common ratio is 2, as each term is double the previous term. This type of series allows us to calculate terms and sums efficiently.
- Formula for the n-th term: In a geometric series, the n-th term can be found using the formula: \(a_n = a_1 \times r^{n-1}\), where \(a_1\) is the first term and \(r\) is the common ratio.
- Sum of the series: The sum of the first \(n\) terms is \(S_n = \frac{a_1 \times (r^n - 1)}{r - 1}\), which makes it easier to calculate large sequences.
Sequence Pattern
Recognizing and understanding sequence patterns is vital in mathematics, especially when dealing with series like this one. In the exercise, identifying the waiting time pattern was crucial to find the total number of payments you would receive.
The sequence starts with a 1-minute wait, doubles to 2 minutes, then 4 minutes, and keeps doubling. This is a classic geometric sequence pattern characterized by its constant ratio between consecutive terms. Knowing this pattern, we can derive any term without listing them all.
- Doubling pattern: Each wait time is twice the previous one: 1, 2, 4, 8, etc.
- Geometric growth: This pattern defines exponential growth, where each step increases at an exponential rate, specifically base 2 in this example.
Series Operations
Series operations involve various mathematical techniques to handle series efficiently. One of the core tasks is the summation of a sequence, as seen in the exercise where we needed to find how much wait time accumulates over various periods.In the given problem, we used a series operation to calculate the sum of the waiting times to ensure it did not exceed a year in minutes. By leveraging the geometric series sum formula, we calculated the highest number of payments, \(n=19\), without surpassing the time limit.
- Accumulating sums: Using formulas like \(S_n = \frac{a_1 \times (r^n - 1)}{r - 1}\) gives precise control over the series' total without manual counting.
- Determining limits: Solving inequalities such as \(2^n - 1 \leq 525,600\) helps determine the maximum feasible length of sequences with given constraints.
Mathematical Modeling
Mathematical modeling is using mathematical structures and techniques to represent and solve real-world problems. In our problem, we modeled a financial sequence with waiting times and payments to determine total earnings over a year.
By representing the wait times as a geometric series, we could mathematically investigate and predict outcomes. This approach allows real-world phenomena, like the distribution of payments over time, to be simplified using mathematical expressions and operations.
- Real-world applications: Simple sequences can model complex scenarios like financial payments or population growth.
- Predictive analysis: By understanding the sequence pattern and utilizing series operations, you can anticipate outcomes.
