Problem 86 - College Algebra with Modeling and Visualization

Question

(Refer to Example \(12 .\) ) Use properties for summation notation to find the sum. $$\sum_{k=1}^{17}(1-4 k)$$

Step-by-Step Solution

Answer

The sum is -595.

Step 1: Break Down the Problem

We need to evaluate the summation \( \sum_{k=1}^{17}(1-4k) \). This means we will calculate the sum of the expression \(1-4k\) as \(k\) goes from 1 to 17.

Step 2: Separate the Summation

Notice the expression inside the summation is \(1 - 4k\). We can use the linearity of summation to separate this into two different summations: \(\sum_{k=1}^{17} 1 - \sum_{k=1}^{17} 4k\).

Step 3: Simplify the First Summation

Calculate \(\sum_{k=1}^{17} 1\). This means we add the number 1, a total of 17 times, so this becomes \(17 \times 1 = 17\).

Step 4: Factor and Simplify the Second Summation

Consider \(\sum_{k=1}^{17} 4k\). We can factor out the 4 as it is a constant, giving us \(4 \sum_{k=1}^{17} k\).

Step 5: Use Arithmetic Series Formula

Now we calculate \(\sum_{k=1}^{17} k\), which is the sum of the first 17 natural numbers. The formula for the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=17\), it is \(\frac{17 \times 18}{2} = 153\).

Step 6: Evaluate the Second Summation

Substitute back into the expression \(4 \sum_{k=1}^{17} k\), which becomes \(4 \times 153 = 612\).

Step 7: Combine the Results

Now combine the results from Step 3 and Step 6: \(17 - 612 = -595\). Therefore, the answer is \(-595\).

Key Concepts

Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence. In an arithmetic sequence, the difference between consecutive terms is a constant, known as the common difference. This makes an arithmetic series simple to manage and predict.

When working with arithmetic series, it is essential to know how to find the sum quickly. The sum of the first \( n \) terms of an arithmetic series can be found using the formula:

• \( S_n = \frac{n}{2} \times (a + l) \)

where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term.

Alternatively, if you only have the first term (\( a \)) and the common difference (\( d \)), you can use:

• \( S_n = \frac{n}{2} \times (2a + (n-1)d) \)

These formulas allow you to quickly calculate the sum without needing to add each term individually.

Linearity of Summation

The linearity of summation is a powerful property that can simplify complex summations. It states that the sum of a sum is the sum of the sums. If you have two functions \( f(k) \) and \( g(k) \), the linearity can be expressed as:

• \( \sum_{k=1}^{n} (f(k) + g(k)) = \sum_{k=1}^{n} f(k) + \sum_{k=1}^{n} g(k) \)

This property allows you to break down a difficult problem into manageable parts.

Additionally, if you have a constant \( c \) that multiplies a function, you can factor it out:

• \( \sum_{k=1}^{n} c \,\cdot\, f(k) = c \,\cdot\, \sum_{k=1}^{n} f(k) \)

This is particularly useful when dealing with terms that are constants or when constants multiply each term inside the summation. This property helps simplify evaluations by separating constants and summing only the variables.

Properties of Summation Notation

Summation notation offers compact and clear expressions for adding sequences of numbers or functions. There are several useful properties to understand:

1. **Reversibility**: Summation can usually be reversed, which means that the series \( \sum_{k=a}^{b} f(k) \) can be rewritten as \( \sum_{k=b}^{a} f(k) \) with appropriately adjusted indices.
2. **Expandability**: An expression like \( \sum_{k=1}^{n} (a_k + b_k) \) can be expanded into \( \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \).
3. **Collapsing**: If the sequence being summed starts and ends with zero, such as \( a_{k+1} - a_k \), the summation will "collapse" or reduce to zero, given the series defined on integral boundaries that naturally lead to cancellation of consecutive terms.

Understanding these properties allows you to manipulate the notation to suit your needs, such as reorganizing, simplifying, or calculating sums more efficiently.