The problem is about evaluating the sum $S_n$ and combining specific terms to find $S_{17} + S_{23} + S_{50}$. Let’s break it down step by step.

Given:

$S_n = 1 - 2 + 3 - 4 + \dots + (-1)^{n-1} n$

This is an alternating sum of integers. The sign alternates between $+$ and $-$, depending on whether the term is odd or even.

Step 1: Find the general formula for $S_n$

We observe:

• If $n$ is odd: The sequence ends with a positive term, and there is an extra positive odd number. Pairing the remaining numbers results in 0. For example: $S_3 = 1 - 2 + 3 = (1 - 2) + 3 = -1 + 3 = 2$ Similarly: $S_5 = 1 - 2 + 3 - 4 + 5 = (1 - 2) + (3 - 4) + 5 = -1 - 1 + 5 = 3$ For odd $n$, the result is: $S_n = \frac{n + 1}{2}.$

• If $n$ is even: The sequence ends with a negative term, and all pairs cancel out to 0. For example: $S_4 = 1 - 2 + 3 - 4 = (1 - 2) + (3 - 4) = -1 - 1 = -2$ Similarly: $S_6 = 1 - 2 + 3 - 4 + 5 - 6 = (1 - 2) + (3 - 4) + (5 - 6) = -1 - 1 - 1 = -3$ For even $n$, the result is: $S_n = -\frac{n}{2}.$

Step 2: Calculate $S_{17}$, $S_{23}$, and $S_{50}$

• $n = 17$ (odd): $S_{17} = \frac{17 + 1}{2} = \frac{18}{2} = 9.$

• $n = 23$ (odd): $S_{23} = \frac{23 + 1}{2} = \frac{24}{2} = 12.$

• $n = 50$ (even): $S_{50} = -\frac{50}{2} = -25.$

Step 3: Find $S_{17} + S_{23} + S_{50}$

$S_{17} + S_{23} + S_{50} = 9 + 12 - 25 = -4.$

Final Answer:

The value of $S_{17} + S_{23} + S_{50}$ is: $\boxed{-4}.$

Would you like me to clarify or expand on any steps? Here are related questions you might consider:

1. How do we derive the alternating series formula?
2. Can we generalize the solution for other values of $n$?
3. How does the pairing technique work for alternating sums?
4. What are some applications of alternating sequences in real life?
5. How do we handle such series in terms of computer algorithms?

Tip: When solving series problems, always check patterns for odd and even cases separately.