Problem 62 - A Graphical Approach to College Algebra

Question

Evaluate each sum. $$\sum_{j=1}^{15}(5 j-9)$$

Step-by-Step Solution

Answer

The sum is 465.

Step 1: Understanding the Problem

We need to evaluate the sum \( \sum_{j=1}^{15} (5j - 9) \). This means we are summing up the expression \( 5j - 9 \) for values of \( j \) ranging from 1 to 15.

Step 2: Expand the Sum

Write out the full expression for the sum: \( (5 \times 1 - 9) + (5 \times 2 - 9) + \cdots + (5 \times 15 - 9) \).

Step 3: Simplify Each Term

Simplify each term: - \( 5 \times 1 - 9 = -4 \) - \( 5 \times 2 - 9 = 1 \) - \( 5 \times 3 - 9 = 6 \)- Continue this pattern up to:- \( 5 \times 15 - 9 = 66 \)

Step 4: Calculate the Partial Sums

Calculate the partial sums: - The terms become \(-4, 1, 6, \ldots, 66\).- Use the arithmetic series sum formula where \( a = -4 \), the first term, \( l = 66 \), the last term, and \( n = 15 \), the number of terms.

Step 5: Use the Arithmetic Sum Formula

The formula for the sum of an arithmetic series is \[ S_n = \frac{n}{2} (a + l) \]Substitute the known values:\[ S_{15} = \frac{15}{2} \times (-4 + 66) \]

Step 6: Perform the Calculations

Calculate the sum using the formula:\[ S_{15} = \frac{15}{2} \times 62 = 15 \times 31 = 465 \]

Step 7: Verify the Result

Ensure the calculated result is correct by checking the calculation steps and verifying that all terms are accounted for.

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a list of numbers in which each term after the first is found by adding a constant to the previous term. This constant is known as the "common difference." In the expression \(5j - 9\), for example, as \(j\) increases, the next term of the sequence can be found by increasing \(j\) by 1, which leads to a constant pattern in the differences between terms.

The common difference can be determined by subtracting any term from the next term. From the term illustrations:

• When \(j = 1\), the term is \(-4\).
• When \(j = 2\), the term is \(1\).

The difference between these consecutive terms is always consistent. For example, \(1 - (-4) = 5\). This tells us the common difference is 5. In an arithmetic sequence, this constancy helps us understand and predict the behavior of the sequence over many terms. Recognizing this sequence aids in solving related series questions efficiently.

Series Calculation

To calculate a series involves the summation of terms from a sequence. When dealing with an arithmetic sequence, the series is simply the sum of the first few or all terms in the sequence.

In this exercise, we are tasked to find the sum of the terms from \(5 \times 1 - 9\) up to \(5 \times 15 - 9\). Each term simplifies to:

• \(5 \times 1 - 9 = -4\)
• \(5 \times 2 - 9 = 1\)
• \(5 \times 3 - 9 = 6\)
• \(\cdots\)
• \(5 \times 15 - 9 = 66\)

All of these calculated terms from \(-4\) to \(66\) can be arranged and summed. Knowing how to organize and calculate this series is essential in solving these types of problems. Each step involves precise simplification and sequencing, preparing us for utilizing more advanced techniques like arithmetic sum formulas.

Summation Techniques

When summing an arithmetic series like \(-4, 1, 6, \ldots, 66\), utilize the formula for the sum of an arithmetic series: \[ S_n = \frac{n}{2} (a + l) \] where:

• \(S_n\) is the sum of the series,
• \(n\) is the total number of terms,
• \(a\) is the first term, and
• \(l\) is the last term.

Replacing these with given values \(a = -4\), \(l = 66\), and \(n = 15\), we substitute into the formula:\[ S_{15} = \frac{15}{2} \times (-4 + 66) \]which calculates to:\[ S_{15} = \frac{15}{2} \times 62 = 15 \times 31 = 465 \]This method capitalizes on the properties of arithmetic series, ensuring accuracy and reducing lengthy calculations. Thoroughly understanding this technique enhances solving other sum-related challenges in mathematics.