In a certain sequence the first term is 1, and the second term is 2. After the second term, each...

1. Translate the problem requirements: We need to understand that this is a recursive sequence where each term after the second equals the previous term minus the term before it. We're looking for how many different possible values exist for the sum of the first n terms for any value of n.
2. Generate the sequence systematically: Calculate the first several terms of the sequence using the given rule to identify any repeating patterns.
3. Calculate cumulative sums: Find the sum of the first n terms for various values of n to see what different sum values are possible.
4. Identify the pattern in sums: Recognize that the sequence is periodic, which means the sums will follow a predictable cycle, limiting the number of possible different sum values.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to do in plain English. We have a sequence that starts with 1, then 2. After that, each new term is found by taking the previous term and subtracting the term before it from that previous term.

So if we call our terms \(\mathrm{a_1, a_2, a_3}\), etc., then:

• \(\mathrm{a_1 = 1}\)
• \(\mathrm{a_2 = 2}\)
• \(\mathrm{a_3 = a_2 - a_1 = 2 - 1 = 1}\)
• \(\mathrm{a_4 = a_3 - a_2 = 1 - 2 = -1}\)
• And so on...

The question asks: "How many different values are possible for the sum of the first n terms?" This means we need to find all the different possible values when we add up the first 1 term, the first 2 terms, the first 3 terms, etc., and see how many distinct sum values we can get.

Process Skill: TRANSLATE - Converting the recursive sequence description into a clear mathematical pattern

2. Generate the sequence systematically

Let's calculate the first several terms to see if we can spot a pattern. Remember, each term after the second equals the previous term minus the term before it.

• \(\mathrm{a_1 = 1}\)
• \(\mathrm{a_2 = 2}\)
• \(\mathrm{a_3 = a_2 - a_1 = 2 - 1 = 1}\)
• \(\mathrm{a_4 = a_3 - a_2 = 1 - 2 = -1}\)
• \(\mathrm{a_5 = a_4 - a_3 = -1 - 1 = -2}\)
• \(\mathrm{a_6 = a_5 - a_4 = -2 - (-1) = -2 + 1 = -1}\)
• \(\mathrm{a_7 = a_6 - a_5 = -1 - (-2) = -1 + 2 = 1}\)
• \(\mathrm{a_8 = a_7 - a_6 = 1 - (-1) = 1 + 1 = 2}\)
• \(\mathrm{a_9 = a_8 - a_7 = 2 - 1 = 1}\)

Look at this! We can see that \(\mathrm{a_7 = 1 = a_1}\), \(\mathrm{a_8 = 2 = a_2}\), and \(\mathrm{a_9 = 1 = a_3}\). This means our sequence repeats every 6 terms!

The sequence is: 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, ...

3. Calculate cumulative sums

Now let's find the sum of the first n terms for various values of n. We'll call these sums \(\mathrm{S_1, S_2, S_3}\), etc.

• \(\mathrm{S_1 = 1}\)
• \(\mathrm{S_2 = 1 + 2 = 3}\)
• \(\mathrm{S_3 = 1 + 2 + 1 = 4}\)
• \(\mathrm{S_4 = 1 + 2 + 1 + (-1) = 3}\)
• \(\mathrm{S_5 = 1 + 2 + 1 + (-1) + (-2) = 1}\)
• \(\mathrm{S_6 = 1 + 2 + 1 + (-1) + (-2) + (-1) = 0}\)

Now, since our sequence repeats every 6 terms, let's see what happens when we continue:

• \(\mathrm{S_7 = S_6 + a_7 = 0 + 1 = 1}\)
• \(\mathrm{S_8 = S_7 + a_8 = 1 + 2 = 3}\)
• \(\mathrm{S_9 = S_8 + a_9 = 3 + 1 = 4}\)
• \(\mathrm{S_{10} = S_9 + a_{10} = 4 + (-1) = 3}\)
• \(\mathrm{S_{11} = S_{10} + a_{11} = 3 + (-2) = 1}\)
• \(\mathrm{S_{12} = S_{11} + a_{12} = 1 + (-1) = 0}\)

4. Identify the pattern in sums

Notice that the sums also follow a repeating pattern! We have:

• \(\mathrm{S_1 = 1, S_7 = 1}\)
• \(\mathrm{S_2 = 3, S_8 = 3}\)
• \(\mathrm{S_3 = 4, S_9 = 4}\)
• \(\mathrm{S_4 = 3, S_{10} = 3}\)
• \(\mathrm{S_5 = 1, S_{11} = 1}\)
• \(\mathrm{S_6 = 0, S_{12} = 0}\)

Since the original sequence repeats every 6 terms, and the sum of one complete cycle \(\mathrm{(a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 0)}\), the pattern of sums will also repeat every 6 terms.

Looking at our calculated sums: 1, 3, 4, 3, 1, 0, and then the pattern repeats.

The distinct sum values are: 0, 1, 3, and 4.

That gives us exactly 4 different possible values for the sum of the first n terms.

Process Skill: INFER - Recognizing that the periodicity of the sequence creates a limited set of possible sum values

4. Final Answer

We found that there are exactly 4 different possible values for the sum of the first n terms: 0, 1, 3, and 4.

The answer is C: 4.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the recursive formula: Students often misinterpret "subtracting from the previous term the term before it" as \(\mathrm{a_3 = a_1 - a_2}\) instead of the correct \(\mathrm{a_3 = a_2 - a_1}\). This fundamental misunderstanding of the sequence generation rule will lead to an entirely different sequence pattern.

2. Not recognizing the need to find periodicity: Students may attempt to calculate individual sums for many values of n without realizing they should look for a repeating pattern. This leads to inefficient problem-solving and potential errors from excessive calculations.

3. Confusing "sum of first n terms" with individual term values: Some students might focus on finding how many different individual term values exist in the sequence rather than understanding that the question asks for different possible cumulative sum values.

Errors while executing the approach

1. Arithmetic errors in sequence generation: When calculating terms like \(\mathrm{a_5 = a_4 - a_3 = -1 - 1 = -2}\) or \(\mathrm{a_6 = a_5 - a_4 = -2 - (-1) = -1}\), students often make sign errors, especially when dealing with negative numbers and subtraction operations.

2. Incorrectly calculating cumulative sums: Students may make addition errors when computing sums like \(\mathrm{S_5 = 1 + 2 + 1 + (-1) + (-2) = 1}\), particularly when mixing positive and negative terms. These errors compound as they continue calculating subsequent sums.

3. Failing to recognize or verify the periodic pattern: Even after calculating several terms, students might not notice that the sequence repeats every 6 terms, or they might incorrectly identify the period length, leading to wrong conclusions about the sum pattern.

Errors while selecting the answer

1. Miscounting distinct sum values: After finding the sum values 0, 1, 3, and 4, students might miscount and select 3 or 5 instead of 4, especially if they mistakenly include or exclude one of the values.

2. Not accounting for all possible sums: Students might stop their analysis too early and miss some possible sum values, or conversely, they might think there are more possible values than actually exist due to incomplete pattern recognition.