Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the value of the indicated sum. $$ \sum_{m=1}^{8}(-1)^{m} 2^{m-2} $$

Step-by-Step Solution

Answer

The sum is \( \frac{85}{2} \).

Step 1: Understand the Summation Notation

The problem requires us to compute \( \sum_{m=1}^{8}(-1)^{m} 2^{m-2} \). This means that we need to calculate the sum by substituting integer values from 1 to 8 into the expression \((-1)^{m} 2^{m-2}\) and adding all those expressions together.

Step 2: Calculate Each Term of the Series

Substitute each integer from 1 to 8 into the expression to obtain each term of the series.- For \( m = 1 \), the term is \((-1)^{1} 2^{1-2} = -\frac{1}{2}\).- For \( m = 2 \), the term is \((-1)^{2} 2^{2-2} = 1\).- For \( m = 3 \), the term is \((-1)^{3} 2^{3-2} = -2\).- For \( m = 4 \), the term is \((-1)^{4} 2^{4-2} = 4\).- For \( m = 5 \), the term is \((-1)^{5} 2^{5-2} = -8\).- For \( m = 6 \), the term is \((-1)^{6} 2^{6-2} = 16\).- For \( m = 7 \), the term is \((-1)^{7} 2^{7-2} = -32\).- For \( m = 8 \), the term is \((-1)^{8} 2^{8-2} = 64\).

Step 3: Sum All Terms Together

Now, add all the calculated terms together: \[-\frac{1}{2} + 1 - 2 + 4 - 8 + 16 - 32 + 64\].First, simplify these terms in groups for easier calculation:1. \(-\frac{1}{2} + 1 = \frac{1}{2} \).2. \( \frac{1}{2} - 2 = -\frac{3}{2} \).3. \(-\frac{3}{2} + 4 = \frac{5}{2} \).4. \( \frac{5}{2} - 8 = -\frac{11}{2} \).5. \(-\frac{11}{2} + 16 = \frac{21}{2} \).6. \( \frac{21}{2} - 32 = -\frac{43}{2} \).7. \(-\frac{43}{2} + 64 = \frac{85}{2} \).

Step 4: Present the Final Result

The calculated sum is \( \frac{85}{2} \), which completes the evaluation of the sum.

Key Concepts

Series Calculation

Series calculation involves finding the sum of a sequence of numbers, often using summation notation. In this problem, we have a series defined by the expression \((-1)^{m} 2^{m-2}\), where \(m\) takes integer values from 1 to 8. Each term of the series is calculated by substituting these integer values into the expression.

• For \(m = 1\) to \(m = 8\), you compute each term individually, giving us a sequence of terms: \(-\frac{1}{2}, 1, -2, 4, -8, 16, -32,\) and \(64\).
• Summing these terms together involves adding both positive and negative values, requiring careful attention to signs and magnitudes.

The key to solving series calculations is to closely follow the sequence of terms and to be methodical in both computing each term and summing them. This process helps ensure all components are correctly calculated and combined.

Alternating Series

An alternating series is a series where the signs of its terms alternate between positive and negative. This is a defining feature of the series \( \sum_{m=1}^{8}(-1)^{m} 2^{m-2} \), where the \((-1)^{m}\) factor causes each term to switch signs.

• Observe how, for odd \(m\), the term \((-1)^{m}\) results in a negative value, while for even \(m\), it results in a positive value.
• This alternating pattern often impacts the convergence or the final summation of the series, as negative and positive terms successively cancel each other out partially.

Understanding the nature of alternating series can simplify the process of summation, especially in identifying patterns or shortcuts to calculate the total sum efficiently.

Exponential Function

The expression \(2^{m-2}\) in our series is an example of an exponential function, where the base (2 in this case) is raised to a power determined by the variable \(m-2\). Exponential functions grow rapidly, especially in the context of powers rising from negative to positive.

• In our series, as \(m\) increases, \(2^{m-2}\) evolves from a fractional power at \(m=1\) to a substantial positive integer at \(m=8\).
• Exponential growth makes earlier terms with smaller exponents relatively smaller in magnitude compared to those with larger exponents.

Grasping the concept of exponential functions helps in understanding how such terms influence the sum heavily as they add significant positive values when \(m\) becomes larger.