The sum of the first 2018 terms of the sequence 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, is equal to

To solve the problem, let's analyze the given sequence pattern:

The sequence is as follows:  
1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, ...

We can see that the sequence consists of the number 1 followed by \( n \) occurrences of the number 2, where \( n \) starts from 1 and increases by 1 after each appearance of the number 1. Let's denote this as follows:

- 1st block: 1, 2 (2 terms)
- 2nd block: 1, 2, 2 (3 terms)
- 3rd block: 1, 2, 2, 2 (4 terms)
- 4th block: 1, 2, 2, 2, 2 (5 terms)

In general, the \( n \)th block contains \( n+1 \) terms.

### Step 1: Find the cumulative sum of terms up to the \( n \)th block.
The number of terms in the first \( n \) blocks is:
\[
2 + 3 + 4 + \cdots + (n+1) = \sum_{k=1}^{n} (k+1) = \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 = \frac{n(n+1)}{2} + n = \frac{n(n+3)}{2}
\]
Thus, the total number of terms up to the \( n \)th block is:
\[
\text{Total terms} = \frac{n(n+3)}{2}
\]

### Step 2: Determine the maximum block \( n \) such that the total number of terms is still less than or equal to 2018.
We solve the inequality:
\[
\frac{n(n+3)}{2} \leq 2018
\]
Multiplying both sides by 2:
\[
n(n+3) \leq 4036
\]
This is a quadratic equation:
\[
n^2 + 3n - 4036 = 0
\]

We solve this using the quadratic formula:
\[
n = \frac{-3 \pm \sqrt{3^2 + 4 \times 4036}}{2}
\]
\[
n = \frac{-3 \pm \sqrt{9 + 16144}}{2} = \frac{-3 \pm \sqrt{16153}}{2}
\]

Approximating \( \sqrt{16153} \approx 127.11 \):
\[
n = \frac{-3 + 127.11}{2} \approx 62.06
\]
Since \( n \) must be an integer, \( n = 62 \).

### Step 3: Calculate the total number of terms and the sum for the first 62 blocks.
For \( n = 62 \):
\[
\text{Total terms} = \frac{62(62+3)}{2} = \frac{62 \times 65}{2} = 2015
\]
There are 2015 terms in the first 62 blocks.

### Step 4: Calculate the sum of these 2015 terms.
For each block \( n \), the sum of terms is:
\[
1 + 2 \times n
\]
So the sum of the first 62 blocks is:
\[
\text{Sum of the first 62 blocks} = \sum_{k=1}^{62} (1 + 2k) = 62 + 2 \sum_{k=1}^{62} k = 62 + 2 \times \frac{62 \times 63}{2} = 62 + 62 \times 63 = 62 + 3906 = 3968
\]

### Step 5: Add the sum of the remaining terms.
There are \( 2018 - 2015 = 3 \) terms remaining, all of which are 2s (from the 63rd block). Therefore, the additional sum is \( 3 \times 2 = 6 \).

### Final Answer:
\[
\text{Total Sum} = 3968 + 6 = 3974
\]

So, the sum of the first 2018 terms of the sequence is **3974**.

Would you like any further details on this solution? Here are five questions related to this problem:

1. How do you derive the formula for the number of terms in the sequence?
2. What method can be used to approximate the roots of a quadratic equation?
3. How would the sum change if the pattern of the sequence were different?
4. Can you explain the summation formula used for the sequence?
5. What are some real-world applications of understanding sequences like this one?

**Tip:** Always double-check the results by re-calculating or testing with smaller examples to ensure the correctness of your solution.

Learn how to calculate the sum of the first 2018 terms of the sequence 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2 using sequence analysis and mathematical formulas. Detailed step-by-step solution provided.

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