It seems like you're asking to evaluate the sum of a series defined by:

$\sum_{k=4}^{17} (3(2k) + 6) - 1250$

Step 1: Simplify the expression inside the sum

We start by simplifying the expression inside the summation:

$3(2k) + 6 = 6k + 6$

Thus, the summation becomes:

$\sum_{k=4}^{17} (6k + 6) - 1250$

Step 2: Split the summation

We can split the sum into two parts:

$\sum_{k=4}^{17} (6k + 6) = \sum_{k=4}^{17} 6k + \sum_{k=4}^{17} 6$

Step 3: Calculate the first sum

The first part involves a sum of multiples of 6:

$\sum_{k=4}^{17} 6k = 6 \sum_{k=4}^{17} k$

The sum of integers from 4 to 17 can be computed using the formula for the sum of an arithmetic series:

$\sum_{k=4}^{17} k = \frac{n}{2} (a + l)$ where:

• $n$ is the number of terms: $n = 17 - 4 + 1 = 14$
• $a$ is the first term: $a = 4$
• $l$ is the last term: $l = 17$

Thus:

$\sum_{k=4}^{17} k = \frac{14}{2} (4 + 17) = 7 \times 21 = 147$

Therefore:

$6 \sum_{k=4}^{17} k = 6 \times 147 = 882$

Step 4: Calculate the second sum

Now, the second part involves a sum of constant values (6):

$\sum_{k=4}^{17} 6 = 6 \times (17 - 4 + 1) = 6 \times 14 = 84$

Step 5: Combine the sums

Now, combine both sums:

$\sum_{k=4}^{17} (6k + 6) = 882 + 84 = 966$

Step 6: Subtract 1250

Finally, subtract 1250 from the result:

$966 - 1250 = -284$

Final Answer:

The value of the summation is:

$\boxed{-284}$

Let me know if you'd like a more detailed breakdown or have any further questions!

Here are some related questions you might find interesting:

1. How do you sum an arithmetic series?
2. What is the difference between a finite sum and an infinite series?
3. How do you evaluate sums with a constant term like $\sum_{k=4}^{17} 6$?
4. What are the general steps to solving summation problems?
5. How does the distributive property apply in summation problems?

Tip: When working with summations, try to split up complex expressions into simpler parts to make the calculation more manageable!