Problem 45 - Algebra and Trigonometry

Question

Find the sum. $$\sum_{k=1}^{5} 2^{k-1}$$

Step-by-Step Solution

Answer

The sum is 31.

Step 1: Understanding the Problem

We need to find the sum of the series given by \( \sum_{k=1}^{5} 2^{k-1} \). This is a summation where the index \( k \) runs from 1 to 5, and the expression \( 2^{k-1} \) is evaluated for each \( k \).

Step 2: Identify the Series Pattern

Let's evaluate \( 2^{k-1} \) for each integer \( k \) from 1 to 5:- When \( k = 1 \): \( 2^{1-1} = 2^0 = 1 \)- When \( k = 2 \): \( 2^{2-1} = 2^1 = 2 \)- When \( k = 3 \): \( 2^{3-1} = 2^2 = 4 \)- When \( k = 4 \): \( 2^{4-1} = 2^3 = 8 \)- When \( k = 5 \): \( 2^{5-1} = 2^4 = 16 \)

Step 3: Calculate the Sum

Now, add up all these values to find the sum:\(1 + 2 + 4 + 8 + 16\)

Step 4: Perform the Addition

Add the values step by step:- First, \( 1 + 2 = 3 \)- Then, \( 3 + 4 = 7 \)- Next, \( 7 + 8 = 15 \)- Finally, \( 15 + 16 = 31 \)

Step 5: Verify the Total

We have calculated each step and verified that the addition is correct. The sum of the series \( \sum_{k=1}^{5} 2^{k-1} \) is 31.

Key Concepts

Exploring Summation Notation

Summation notation is a convenient and compact way to write the sum of a sequence of numbers. It utilizes the Greek letter Sigma (∑) to indicate 'sum'. In our exercise, \[ \sum_{k=1}^{5} 2^{k-1} \] the Sigma symbol expresses that we need to calculate the total sum starting from the lower index, \( k = 1 \), to the upper index, \( k = 5 \). This method simplifies the representation of summing sequences, especially useful for expressing long or complex series.

• The numbers below and above the Sigma sign are called the limits; they tell you where to start and stop.
• The expression following the Sigma (in this case \( 2^{k-1} \)) determines each term in the series.
• Each \( k \) value from the lower to upper limit is substituted into the expression to calculate individual terms.

Mastering summation notation is crucial for dealing with series in both arithmetic and geometric contexts, because of its power to succinctly express what would otherwise be bulky, repetitive calculations.

Understanding Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent. In our exercise, the base is 2 and changes occur through different exponents as determined by the index \( k-1 \). Each term in the sequence is calculated as \( 2^{k-1} \), which represents exponential growth, meaning there is a consistent rate of change from one term to the next compared to a linear increase.

• An exponential function like \( 2^{k-1} \) grows rapidly as k increases. Small changes in \( k \) lead to large changes in the value.
• Exponential functions are common in series because they model phenomena with rapid growth, such as populations, radioactive decay, or in this case, geometric series.

In this exercise, recognizing the exponential nature of the terms helps to predict the sequence's behavior and confirms calculations, like how the series progresses by multiplying the previous term by 2.

Recognizing Sequence Patterns

Identifying patterns in sequences is an essential skill in understanding and predicting series behavior. In the exercise given, we see that the sequence generated by the terms \( 2^{k-1} \) gives us a pattern of doubling the preceding term: 1, 2, 4, 8, 16. This pattern illustrates a geometric sequence, where each term is obtained by multiplying the previous term by a common ratio (in this case, 2).

• A geometric sequence is one where the ratio between consecutive terms is constant.
• This particular pattern, because of its geometric nature, results in fast progression in values.
• Identifying the pattern can also help verify calculations; if a term doesn't fit, it signals a computation error.

Developing the knack for spotting sequence patterns is valuable in math as it aids in simplifying problem-solving processes and boosts your ability to work through more complex calculations.