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.