Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, there's a pattern here with alternating signs. Let me write out the first few terms to get a better sense. 

So, starting with 1, then subtract 2, add 3, subtract 4, add 5, subtract 6, and so on until we subtract 100. It looks like the signs alternate between positive and negative with each consecutive term. 

Maybe I can group the terms in pairs. Let's try that. The first pair is 1 - 2, which equals -1. The next pair is 3 - 4, which is also -1. Then 5 - 6 is -1 again. Wait, so each pair of consecutive numbers results in -1. 

If that's the case, then how many such pairs are there? The sequence goes up to 100, which is the 100th term. Since each pair consists of two terms, the total number of pairs would be 100 divided by 2, which is 50. So there are 50 pairs, each summing to -1. 

Multiplying the number of pairs by the sum of each pair gives 50 * (-1) = -50. 

But wait, let me double-check this. Let's test with a smaller example. Suppose we have 1 - 2 + 3 - 4. Grouping them into pairs: (1 - 2) + (3 - 4) = (-1) + (-1) = -2. Which is indeed 50 pairs each giving -1, but in this case, with 4 terms, it's 2 pairs. So 2 * (-1) = -2, which matches. 

Another check: 1 - 2 + 3 - 4 + 5 - 6. That's three pairs: (1-2) + (3-4) + (5-6) = -1 -1 -1 = -3. Which is 3 pairs * (-1) = -3. So this seems to hold. 

Therefore, applying this to the original problem with 100 terms, there are 50 pairs, each summing to -1, leading to a total of -50. 

But let me think again if there's another way to approach this. Maybe using the formula for an arithmetic series with alternating signs. The general formula for such a series is a bit different. 

Wait, the series is 1 - 2 + 3 - 4 + ... + (-1)^(n-1)*n. For n terms. But in our case, n is 100. 

Alternatively, we can consider the sum S = 1 - 2 + 3 - 4 + ... + 99 - 100. 

Another approach: Let's write the sum as the sum of odd terms minus the sum of even terms. 

Odd terms are 1, 3, 5, ..., 99. How many odd terms are there? From 1 to 99, stepping by 2. The number of terms is (99 - 1)/2 + 1 = 50 terms. 

Similarly, even terms are 2, 4, 6, ..., 100. There are also 50 terms. 

Sum of odd terms: The sum of the first 50 odd numbers. The formula for the sum of the first n odd numbers is n^2. So 50^2 = 2500. 

Sum of even terms: The sum of the first 50 even numbers. The formula for that is n*(n + 1). So 50*51 = 2550. 

Therefore, the total sum S = sum of odd terms - sum of even terms = 2500 - 2550 = -50. 

Same result as before. So this confirms the answer. 

Alternatively, using another method: Let's consider the sum S = 1 - 2 + 3 - 4 + ... + 99 - 100. 

If we add 100 to both sides, we get S + 100 = 1 - 2 + 3 - 4 + ... + 99. Then, subtract the original sum S: (S + 100) - S = (1 - 2 + 3 - 4 + ... + 99) - (1 - 2 + 3 - 4 + ... + 99 - 100). 

This simplifies to 100 = 100. Wait, that doesn't help. Maybe another approach. 

Alternatively, consider the sum as alternating sum. 

Another way: Let's pair the terms as follows: (1 + (-2)) + (3 + (-4)) + ... + (99 + (-100)). Each pair is (2k - 1) - 2k = -1. There are 50 such pairs, so total sum is 50*(-1) = -50. 

Same result again. 

So all methods converge to -50. Therefore, the answer is -50. 

I think that's solid. All different approaches lead to the same result, so I can be confident that the answer is correct.
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