Problem 47
Question
Find the sum. $$\sum_{k=1}^{5} 2^{k-1}$$
Step-by-Step Solution
Verified Answer
The sum is 31.
1Step 1: 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.
2Step 2: 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 \(2^{5-1} = 2^{4} = 16\).
3Step 3: 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: - \(1 + 2 = 3\)- \(3 + 4 = 7\)- \(7 + 8 = 15\)- \(15 + 16 = 31\)
4Step 4: 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.
Key Concepts
Evaluating SequencesPower of TwoAdding Series Terms
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, \( 15 + 16 = 31 \).
Other exercises in this chapter
Problem 47
Simplify using the Binomial Theorem. $$\frac{(x+h)^{3}-x^{3}}{h}$$
View solution Problem 47
Find the partial sum \(S_{n}\) of the arithmetic sequence that satisfies the given conditions. $$a_{1}=55, d=12, n=10$$
View solution Problem 48
Simplify using the Binomial Theorem. $$\frac{(x+h)^{4}-x^{4}}{h}$$
View solution Problem 48
Find the sum. $$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\dots-\frac{1}{512}$$
View solution