1001 ∑ (3k-1) K=1 Use wolfram Alpha on question 1-4 now solve this

The summation you've provided is:

\[
\sum_{k=1}^{1001} (3k - 1)
\]

Let's break down how to solve it.

1. **Separate the Sum**: You can separate this sum into two parts:
   \[
   \sum_{k=1}^{1001} (3k - 1) = \sum_{k=1}^{1001} 3k - \sum_{k=1}^{1001} 1
   \]

2. **Factor Out Constants**:
   - For the first part: \( \sum_{k=1}^{1001} 3k = 3 \sum_{k=1}^{1001} k \)
   - For the second part: \( \sum_{k=1}^{1001} 1 = 1001 \), since summing 1 over 1001 terms simply gives 1001.

3. **Evaluate the Sum of \( k \)**:
   Using the formula for the sum of the first \( n \) integers, \( \sum_{k=1}^n k = \frac{n(n+1)}{2} \), we find:
   \[
   \sum_{k=1}^{1001} k = \frac{1001 \cdot 1002}{2} = 501501
   \]

4. **Substitute Back**:
   Now, calculate each term:
   - \( 3 \sum_{k=1}^{1001} k = 3 \times 501501 = 1504503 \)
   - \( \sum_{k=1}^{1001} 1 = 1001 \)

5. **Combine the Results**:
   \[
   \sum_{k=1}^{1001} (3k - 1) = 1504503 - 1001 = 1503502
   \]

### Final Answer:
\[
\sum_{k=1}^{1001} (3k - 1) = 1503502
\]

Would you like further explanation on any step, or assistance with questions 2-4?

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Here are five related questions to expand on this topic:

1. How does the formula for summing an arithmetic sequence differ from this summation approach?
2. What happens if we change the upper limit from 1001 to a different number?
3. How would the answer differ if we had a different constant multiplier in \(3k - 1\)?
4. Can we apply similar steps if \(3k - 1\) was instead \(3k^2 - 1\)?
5. How would the approach change if the terms alternated signs, such as \(3k - 1\) for even \(k\) and \(1 - 3k\) for odd \(k\)?

**Tip:** For large summations, breaking the expression down and factoring constants often simplifies the calculation.

This solution walks through calculating the summation of (3k - 1) from k=1 to 1001 using arithmetic series principles. It provides a detailed step-by-step breakdown, suitable for high school students studying summations and series.

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