What is the sum of odd integers from 35 to 85, inclusive?

1. Translate the problem requirements: We need to find the sum of all odd integers starting from 35 and ending at 85, including both endpoints. This means we're looking for: \(35 + 37 + 39 + \ldots + 83 + 85\).
2. Identify the arithmetic sequence pattern: Recognize that consecutive odd numbers form an arithmetic sequence with a common difference of 2, and determine the first term, last term, and count of terms.
3. Count the number of terms: Use the relationship between first term, last term, and common difference to find how many odd numbers are in our range.
4. Apply the arithmetic series sum formula: Use the formula \(\mathrm{Sum} = (\mathrm{number\ of\ terms}) \times (\mathrm{first\ term} + \mathrm{last\ term}) \div 2\) to calculate the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding exactly what we're looking for. We need to find the sum of all odd integers from 35 to 85, inclusive.

The word "inclusive" means we include both 35 and 85 in our calculation. So we're adding up: \(35 + 37 + 39 + 41 + \ldots + 81 + 83 + 85\).

Notice that we're only looking at odd numbers in this range. We skip all the even numbers like 36, 38, 40, etc.

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical requirement

2. Identify the arithmetic sequence pattern

Let's look at how odd numbers behave when we list them out:

\(35, 37, 39, 41, 43, \ldots\)

Do you notice the pattern? Each odd number is exactly 2 more than the previous odd number. This is what mathematicians call an "arithmetic sequence" - a sequence where we add the same amount each time to get the next term.

In our case:

• First term = 35
• Last term = 85
• Common difference = 2 (the amount we add each time)

This pattern makes our calculation much easier because arithmetic sequences have a simple formula for finding their sum.

3. Count the number of terms

Before we can find the sum, we need to know how many odd numbers are between 35 and 85, inclusive.

Here's a logical way to think about it: If we have an arithmetic sequence, we can find the number of terms using the relationship between the first term, last term, and common difference.

The formula is: \(\mathrm{Number\ of\ terms} = (\mathrm{Last\ term} - \mathrm{First\ term}) \div \mathrm{Common\ difference} + 1\)

Let's substitute our values:

\(\mathrm{Number\ of\ terms} = (85 - 35) \div 2 + 1\)

\(\mathrm{Number\ of\ terms} = 50 \div 2 + 1\)

\(\mathrm{Number\ of\ terms} = 25 + 1 = 26\)

So there are 26 odd numbers from 35 to 85, inclusive.

Process Skill: INFER - Drawing the non-obvious conclusion that we need to count terms before finding the sum

4. Apply the arithmetic series sum formula

Now we can find the sum using the arithmetic series formula. Here's the intuitive way to think about it:

Imagine pairing up terms from opposite ends: \((35 + 85), (37 + 83), (39 + 81)\), and so on. Each pair adds up to 120!

The arithmetic series sum formula captures this pattern: \(\mathrm{Sum} = (\mathrm{Number\ of\ terms}) \times (\mathrm{First\ term} + \mathrm{Last\ term}) \div 2\)

Let's substitute our values:

\(\mathrm{Sum} = 26 \times (35 + 85) \div 2\)

\(\mathrm{Sum} = 26 \times 120 \div 2\)

\(\mathrm{Sum} = 26 \times 60\)

\(\mathrm{Sum} = 1,560\)

4. Final Answer

The sum of odd integers from 35 to 85, inclusive, is 1,560.

This matches answer choice A) 1,560.

Quick verification: Our answer makes sense because we're adding 26 numbers that average around 60 (the middle value between 35 and 85), so \(26 \times 60 = 1,560\) is reasonable.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "inclusive" constraint: Students often forget that "inclusive" means both 35 and 85 must be included in the sum. They might exclude one or both endpoints, leading to an incorrect count of terms and wrong final answer.

2. Including even numbers in the sequence: The problem specifically asks for "odd integers" but students may mistakenly include all integers from 35 to 85. This would lead them to use the wrong formula or count, significantly inflating their final answer.

3. Not recognizing the arithmetic sequence pattern: Students might attempt to manually add all odd numbers \((35+37+39+\ldots)\) instead of recognizing this as an arithmetic sequence with first term 35, last term 85, and common difference 2. This makes the problem much more time-consuming and error-prone.

Errors while executing the approach

1. Incorrect term counting formula: When using the formula "\(\mathrm{Number\ of\ terms} = (\mathrm{Last\ term} - \mathrm{First\ term}) \div \mathrm{Common\ difference} + 1\)", students frequently forget to add the "+1" at the end. This would give them 25 terms instead of 26, leading to a final answer of 1,500 instead of 1,560.

2. Arithmetic errors in the sum formula: Students might correctly identify all components but make calculation mistakes. For example, they might calculate \(26 \times 120 \div 2\) as \(26 \times 120 = 3,120\), then forget to divide by 2, or incorrectly perform the division.

3. Using wrong sum formula: Students might confuse the arithmetic series sum formula and use \(\mathrm{Sum} = n \times (\mathrm{first\ term})\) instead of \(\mathrm{Sum} = n \times (\mathrm{first\ term} + \mathrm{last\ term}) \div 2\), leading to a significantly incorrect result.

Errors while selecting the answer

No likely faltering points - once students have correctly calculated 1,560, it directly matches answer choice A with no additional interpretation needed.