<Question>
Compute: $1-2+3-4+5- \dots +99-100$.
</Question>

<think>
<Plan_1>
Okay, so I need to compute the sum of the series 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let me think about how to approach this. 

First, I notice that the series alternates between adding an odd number and subtracting an even number. So, it's like 1 minus 2 plus 3 minus 4 and so on, all the way up to 99 minus 100. Since the last term is 100, which is even, it's subtracted. So the sequence goes up to 100, but the last operation is subtraction. 

Let me write out the first few terms to see the pattern:
1 - 2 + 3 - 4 + 5 - 6 + ... + 99 - 100.

Now, I need to figure out how many terms there are. Since we're going from 1 to 100, that's 100 terms. But since the signs alternate starting with a positive for 1, then negative for 2, etc. , the pattern is +, -, +, -, ...., so each pair of terms is a positive odd number and a negative even number. 

Wait, so maybe I can group the terms in pairs. Let's see:

(1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100).

If I do that, each pair is (odd - even). Let's compute one pair: 1 - 2 = -1, 3 - 4 = -1, 5 - 6 = -1, etc. So each pair sums to -1. Now, how many such pairs are there?

Since there are 100 terms total, and each pair has 2 terms, there are 100 / 2 = 50 pairs. So if each pair is -1, then the total sum would be 50 * (-1) = -50.

But wait, let me check with a smaller example to see if this logic holds. Let's take n=2: 1 - 2 = -1.

<prompt_1.1> Check the logic of pairing terms using a smaller example where n=2. </prompt_1.1>

Another example: n=4. 1 - 2 + 3 - 4 = (1-2) + (3-4) = (-1) + (-1) = -2. Which is -4/2 = -2. So that works.

Similarly, n=6: 1-2+3-4+5-6 = (1-2)+(3-4)+(5-6) = -3, which is -6/2 = -3.

So the formula seems to be that for n terms, where n is even, the sum is -n/2.

<prompt_1.2> Derive a general formula for the sum based on the pattern observed. </prompt_1.2>

But let me make sure there isn't another way to think about this. Maybe using arithmetic series or something else. 

Alternatively, let's consider the series as the sum of all odd numbers from 1 to 99 minus the sum of all even numbers from 2 to 100. 

Let's compute that. 

First, the sum of the odd numbers from 1 to 99. How many odd numbers are there between 1 and 99 inclusive? Since 99 is odd, the count is (99 - 1)/2 + 1 = 49 + 1 = 50 terms. The sum of the first m odd numbers is m^2.

<prompt_1.3> Calculate the sum of odd numbers using the formula for the sum of the first m odd numbers. </prompt_1.3>

Now, the sum of even numbers from 2 to 100. There are 100/2 = 50 terms. The sum of an arithmetic series is n/2*(first term + last term). So, 50/2*(2 + 100) = 25*102 = 2550.

<prompt_1.4> Calculate the sum of even numbers and find the final result by subtracting from the sum of odd numbers. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
According to the pairing method, it would be (1-2) = -1, which is correct. For n=2, the result is -1, which is -2/2 = -1. So that works.
</execution_1.1>

<execution_1.2>
So if the last term is 100, which is even, then the number of pairs is 100/2 = 50, each pair summing to -1, so total sum is -50. 

But wait, let me confirm again with another example. Let's take n=6: 1-2+3-4+5-6. The pairs are (1-2), (3-4), (5-6). Each is -1, so 3*(-1) = -3, which is -6/2 = -3.
</execution_1.2>

<execution_1.3>
Wait, but that formula is for the sum 1 + 3 + 5 + ... + (2m-1) = m². Here, our last term is 99. Let's see, 2m - 1 = 99 => m = 50. So the sum is 50² = 2500.
</execution_1.3>

<execution_1.4>
So, sum of odds is 50^2 = 2500. Then total sum is 2500 - 2550 = -50.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.4:
Therefore, the answer is -50. 

But just to be thorough, let me write out the steps again:

<prompt_2.1> Reiterate the steps and calculations to confirm the solution. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
1. Identify the pattern: alternating addition and subtraction starting with +1, -2, +3, -4, ...., +99, -100.

2. Group the terms into pairs: (1-2), (3-4), ...., (99-100). Each pair has two terms.

3. Each pair sums to -1. Number of pairs: 100 terms / 2 = 50 pairs.

4. Total sum: 50 * (-1) = -50.

5. Alternatively, separate the series into sum of odd numbers (positive) and sum of even numbers (negative):

Sum = (1 + 3 + 5 + ... + 99) - (2 + 4 + 6 + ... + 100).

Sum of odds: 50 terms, sum = 50² = 2500.

Sum of evens: 50 terms, sum = 50/2*(2 + 100) = 25*102 = 2550.

Total sum: 2500 - 2550 = -50.
</execution_2.1>
</Execution_2>

<Final_answer>
Therefore, the final answer is $\boxed{-50}$.
</Final_answer>
</think>