I currently don’t have direct access to Wolfram Alpha, but I can help solve this problem using my own computational abilities.

The sum in question 1 is:

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

Let's break it down:

1. We can separate the terms inside the summation:

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

2. Now we handle each part separately.

   • First Part: $\sum_{k=1}^{1001} 3k$

     Since 3 is a constant multiplier, it can be taken outside the summation:

     $3 \sum_{k=1}^{1001} k$

     The formula for the sum of the first $n$ natural numbers is $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.

     So, substituting $n = 1001$:

     $\sum_{k=1}^{1001} k = \frac{1001 \cdot 1002}{2} = 501501$

     Therefore,

     $3 \sum_{k=1}^{1001} k = 3 \cdot 501501 = 1504503$

   • Second Part: $\sum_{k=1}^{1001} 1$

     This is simply adding 1 a total of 1001 times, so:

     $\sum_{k=1}^{1001} 1 = 1001$

3. Now, combining both results:

   $\sum_{k=1}^{1001} (3k - 1) = 1504503 - 1001 = 1503502$

So, the answer to the sum is:

$\sum_{k=1}^{1001} (3k - 1) = 1503502$

If you need further clarification or additional details, let me know!

Here are 5 related questions to consider:

1. How would the solution change if we summed from $k = 1$ to $k = n$ instead of 1001?
2. What if the formula inside the summation were different, such as $4k + 2$?
3. How can we apply summation properties to simplify expressions like this further?
4. What is the general formula for the sum of an arithmetic series?
5. How would this approach change if $k$ had different limits, like starting from $k = 0$ instead of $k = 1$?

Tip: When dealing with summations, always look for ways to separate constants and simplify by using known summation formulas for basic series, such as $\sum k$ or $\sum 1$.