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Magazine Chapter 12: Problem 47
Find the sum. $$\sum_{k=1}^{5} 2^{k-1}$$
Short Answer
Expert verified
The sum is 31.
Step by step solution
01
Understanding the Summation Notation
The given exercise asks us to find the sum of the expression \( \sum_{k=1}^{5} 2^{k-1} \). This notation indicates that we need to evaluate \(2^{k-1}\) for each integer value of \(k\) from 1 to 5, and then add all these evaluated terms together.
02
Evaluating Terms of the Sequence
We systematically calculate each term: - When \(k=1\), the term is \(2^{1-1} = 2^{0} = 1\).- When \(k=2\), the term is \(2^{2-1} = 2^{1} = 2\).- When \(k=3\), the term is \(2^{3-1} = 2^{2} = 4\).- When \(k=4\), the term is \(2^{4-1} = 2^{3} = 8\).- When \(k=5\), the term is \(2^{5-1} = 2^{4} = 16\).
03
Adding the Evaluated Terms
Now, we add up all the terms calculated in the previous step:\[1 + 2 + 4 + 8 + 16\]. Carrying out the addition step-by-step: - \(1 + 2 = 3\)- \(3 + 4 = 7\)- \(7 + 8 = 15\)- \(15 + 16 = 31\)
04
Conclusion
The sum of the series \( \sum_{k=1}^{5} 2^{k-1} \) is 31. We have added together all the terms to find the total sum given in the problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Evaluating Sequences
Evaluating sequences is a fundamental concept in mathematics where we need to understand how to approach a list of numbers or terms systematically. In the context of our exercise, a sequence is formed by evaluating the expression \( 2^{k-1} \) for different integer values of \( k \) within a specified range. Each time \( k \) changes, the expression produces a different number.
For our specific exercise, we start with \( k = 1 \) and go up to \( k = 5 \), evaluating \( 2^{k-1} \) for each value:
For our specific exercise, we start with \( k = 1 \) and go up to \( k = 5 \), evaluating \( 2^{k-1} \) for each value:
- For \( k = 1 \), we have \( 2^{1-1} = 2^0 = 1 \).
- For \( k = 2 \), we find \( 2^{2-1} = 2^1 = 2 \).
- \( k = 3 \) yields \( 2^{3-1} = 2^2 = 4 \).
- With \( k = 4 \), the term becomes \( 2^{4-1} = 2^3 = 8 \).
- Finally, \( k = 5 \) results in \( 2^{5-1} = 2^4 = 16 \).
Power of Two
The sequence we are dealing with in this exercise is constructed using powers of two, which is a common mathematical topic. The expression \( 2^{k-1} \) shows how powers of two originate from a base of 2 raised to a particular exponent.
Powers of two are significant because they frequently appear in various areas, such as computer science, to represent binary numbers or memory storage. In the step-by-step breakdown:
Powers of two are significant because they frequently appear in various areas, such as computer science, to represent binary numbers or memory storage. In the step-by-step breakdown:
- \( 2^0 = 1 \) is the start point, denoting a single unit.
- \( 2^1 = 2 \) allows us to double the previous amount.
- \( 2^2 = 4 \) continues this pattern by doubling again.
- Similarly, \( 2^3 = 8 \) and \( 2^4 = 16 \) keep multiplying the previous result by 2.
Adding Series Terms
In mathematics, adding series terms is crucial for determining the total value of a sequence of numbers. Once we have evaluated each term in our sequence, the next step is summing them up, which transforms separate entities into a cumulative total.
For this exercise, after determining that our terms are 1, 2, 4, 8, and 16, we proceed to add them:
For this exercise, after determining that our terms are 1, 2, 4, 8, and 16, we proceed to add them:
- Adding the first two terms: \( 1 + 2 = 3 \).
- Then, \( 3 + 4 = 7 \).
- Next, \( 7 + 8 = 15 \).
- Finally, combining with the last term, \( 15 + 16 = 31 \).
