If the sum of the even integers between 1 and n is 79 * 80, where n is an odd...

1. Translate the problem requirements: We need to find an odd integer n where the sum of all even integers from 2 to n-1 equals \(79 \times 80 = 6320\)
2. Identify the even integers in our range: Since n is odd and we want even integers between 1 and n, we're looking at 2, 4, 6, ..., up to the largest even integer less than n
3. Express the sum using arithmetic sequence principles: The even integers form the sequence 2, 4, 6, ..., (n-1), which is an arithmetic sequence where we can find the sum
4. Set up and solve the equation: Use the relationship between the number of terms and the sum to find n

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking us to find.

We're told that n is an odd integer, and we need to find the sum of all even integers "between 1 and n". The key word here is "between" - this means we don't include 1 or n themselves, just the numbers in the middle.

Since 1 is odd and n is odd, the even integers between them would start at 2 and go up to the largest even number that's still less than n. If n is odd, then n-1 is even, so our even integers are: \(2, 4, 6, 8, \ldots, (\mathrm{n}-1)\).

We're told this sum equals \(79 \times 80\). Let's calculate: \(79 \times 80 = 6,320\).

So we need to find the odd integer n where the sum \(2 + 4 + 6 + \ldots + (\mathrm{n}-1) = 6,320\).

Process Skill: TRANSLATE - Converting the phrase "even integers between 1 and n" into the concrete sequence \(2, 4, 6, \ldots, (\mathrm{n}-1)\)

2. Identify the even integers in our range

Now let's think about what this sequence looks like more clearly.

If n is odd, then n-1 is even. So our sequence of even integers is:
\(2, 4, 6, 8, \ldots, (\mathrm{n}-1)\)

How many even integers are there in this sequence?

Well, the even integers from 2 up to any even number k are just: \(2, 4, 6, \ldots, \mathrm{k}\)
The number of terms in this sequence is \(\mathrm{k} \div 2\).

Since our last term is \((\mathrm{n}-1)\), and n-1 is even, we have \((\mathrm{n}-1) \div 2\) terms total.

For example, if n = 11, then n-1 = 10, and our even integers are 2, 4, 6, 8, 10. That's \(10 \div 2 = 5\) terms.

3. Express the sum using arithmetic sequence principles

Now we can find the sum using a simple principle: when you have evenly spaced numbers, the sum equals the average times the number of terms.

Our sequence is: \(2, 4, 6, \ldots, (\mathrm{n}-1)\)

• First term = 2
• Last term = n-1
• Number of terms = \((\mathrm{n}-1) \div 2\)

The average of the first and last term is: \((2 + (\mathrm{n}-1)) \div 2 = (\mathrm{n}+1) \div 2\)

So the sum is: (average) × (number of terms) = \([(\mathrm{n}+1) \div 2] \times [(\mathrm{n}-1) \div 2]\)

This gives us: Sum = \((\mathrm{n}+1)(\mathrm{n}-1) \div 4 = (\mathrm{n}^2 - 1) \div 4\)

4. Set up and solve the equation

Now we can set up our equation. We know the sum equals 6,320:

\((\mathrm{n}^2 - 1) \div 4 = 6,320\)

Multiplying both sides by 4:
\(\mathrm{n}^2 - 1 = 25,280\)

Adding 1 to both sides:
\(\mathrm{n}^2 = 25,281\)

Taking the square root:
\(\mathrm{n} = \sqrt{25,281} = 159\)

Let's verify this makes sense: if n = 159, then our even integers are 2, 4, 6, ..., 158.

• Number of terms: \(158 \div 2 = 79\)
• Sum = \((2 + 158) \times 79 \div 2 = 160 \times 79 \div 2 = 80 \times 79 = 6,320\) ✓

Notice how this matches our given expression \(79 \times 80 = 6,320\)!

Process Skill: INFER - Recognizing that \(79 \times 80\) in the problem statement actually represents the number of terms (79) times the average (80)

Final Answer

n = 159

The answer is D) 159.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "between 1 and n"

Students often include the endpoints when they see "between 1 and n", thinking the sequence should be 2, 4, 6, ..., n instead of 2, 4, 6, ..., (n-1). This happens because they don't carefully consider that "between" excludes the boundaries, and since n is odd, it wouldn't be part of the even integer sequence anyway.

2. Forgetting the constraint that n is odd

Students may overlook that n must be odd, which is crucial for determining that the last even integer in the sequence is (n-1). Without this constraint, they might assume n could be even and set up their sequence incorrectly.

3. Incorrectly identifying the arithmetic sequence formula

Some students might try to use the standard arithmetic sequence sum formula \(\mathrm{S} = \mathrm{n}/2[2\mathrm{a} + (\mathrm{n}-1)\mathrm{d}]\) without properly identifying that their sequence is 2, 4, 6, ... which has first term a = 2 and common difference d = 2. They might confuse this with the general consecutive integer formula.

Errors while executing the approach

1. Arithmetic errors in calculating \(79 \times 80\)

Students may incorrectly calculate \(79 \times 80\), getting values like 6,240 or 6,400 instead of 6,320. This computational error would lead them to solve for the wrong value of \(\mathrm{n}^2\) and arrive at an incorrect final answer.

2. Making algebraic mistakes when solving \(\mathrm{n}^2 - 1 = 25,280\)

When solving the equation, students might forget to add 1 to both sides correctly, getting \(\mathrm{n}^2 = 25,280\) instead of \(\mathrm{n}^2 = 25,281\). Alternatively, they might make errors when taking the square root, especially if they don't recognize that 25,281 is a perfect square.

3. Errors in the sum formula application

Students may incorrectly apply the formula \((\mathrm{n}+1)(\mathrm{n}-1)/4\), either by making sign errors in the difference of squares or by incorrectly calculating the number of terms as \((\mathrm{n}-1)/2\). These errors in the fundamental setup will propagate through to the final calculation.

Errors while selecting the answer

1. Selecting n-1 instead of n

After correctly calculating that the last even integer in their sequence is 158, students might mistakenly think this is the answer, selecting 158 or the closest option rather than recognizing that n = 159. This happens because they confuse the last term of their sequence with the value of n itself.

2. Not verifying that the chosen answer satisfies all constraints

Students may arrive at a numerical answer but fail to check that it's odd (as required by the problem) or that it actually produces the given sum. Without verification, they might select an even number from the choices if their calculation was slightly off.