Explanation:
Okay, so I need to find the length of DE in this diagram. Let me try to visualize it first. There's a right triangle DEF, right? Because there's a right angle mark at E, between D, E, and F. So, the right angle is at E. Let me confirm that from the Asymptote code. The Asymptote code defines points D, E, F. F is at (0,0), D is at (sqrt(51),7), and E is at (0,7). Then it draws D connected to E, E connected to F, and F connected back to D. The right angle mark is at D, E, F, so the right angle is at E. So triangle DEF is a right triangle with right angle at E.

Given that sin D = 0.7, and we need to find DE. Let me recall that in a right triangle, the sine of an angle is opposite over hypotenuse. So angle D is at point D, so let me think about the sides relative to angle D.

In triangle DEF, angle D is at vertex D. The sides of the triangle are DE, DF, and EF. Since the right angle is at E, the hypotenuse would be DF. Wait, because in a right triangle, the hypotenuse is opposite the right angle. So if the right angle is at E, then the hypotenuse is DF. Therefore, angle D is one of the acute angles.

So, sin D would be opposite over hypotenuse. The side opposite angle D is EF, and the hypotenuse is DF. Therefore, sin D = EF / DF = 0.7.

But wait, let me make sure. Let me label the triangle properly. Let me sketch mentally: point E is at (0,7), F is at (0,0), so EF is a vertical line segment from (0,0) to (0,7), so that's length 7. Then D is at (sqrt(51),7), so that's a point with the same y-coordinate as E, so DE is a horizontal line from E (0,7) to D (sqrt(51),7), so DE is horizontal. Then DF connects D (sqrt(51),7) to F (0,0). So triangle DEF has right angle at E, with legs EF (vertical) and DE (horizontal), and hypotenuse DF.

Therefore, angle D is at point D, so looking at angle D, the sides adjacent and opposite. Let me think again. For angle D, the sides are: the side opposite to angle D is EF, which is 7 units long. The hypotenuse is DF. The side adjacent to angle D is DE.

Therefore, sin D = opposite / hypotenuse = EF / DF = 7 / DF = 0.7. Therefore, 7 / DF = 0.7, so DF = 7 / 0.7 = 10. Therefore, DF is 10.

But wait, the problem is asking for DE. So maybe I need to find DE? But if I can find DF, which is 10, and I know EF is 7, then since DEF is a right triangle, by Pythagoras, DE^2 + EF^2 = DF^2. Therefore, DE^2 + 7^2 = 10^2. Therefore, DE^2 + 49 = 100. Therefore, DE^2 = 51. Therefore, DE = sqrt(51).

Wait, but the coordinates of D are given as (sqrt(51),7). Since E is at (0,7), then DE is the horizontal distance between (sqrt(51),7) and (0,7), which is sqrt(51) - 0 = sqrt(51). So DE is sqrt(51). Therefore, the answer is sqrt(51).

But let me check if this is consistent with the given sin D = 0.7. Let me verify. If DE is sqrt(51), EF is 7, DF is 10. Then sin D is EF / DF = 7 / 10 = 0.7, which matches the given value. So that seems correct.

But wait, the problem says "Suppose sin D = 0.7 in the diagram below." But in the Asymptote code, they have coordinates for D, E, F. So maybe they are giving coordinates, but the problem is just asking for DE? But according to the coordinates, DE is sqrt(51) - 0 = sqrt(51). But maybe the problem is not using the coordinates? Wait, the Asymptote code is just for the diagram, but the problem says "Suppose sin D = 0.7 in the diagram below." So maybe the diagram is just a generic diagram, and the coordinates are just for drawing? Wait, but the Asymptote code defines specific coordinates. Let me check again.

Wait, the Asymptote code defines F at (0,0), D at (sqrt(51),7), E at (0,7). So in that diagram, DE is sqrt(51), EF is 7, DF is sqrt( (sqrt(51))^2 + 7^2 ) = sqrt(51 + 49) = sqrt(100) = 10. Therefore, sin D is EF / DF = 7 / 10 = 0.7, which matches the given value. Therefore, in this diagram, DE is sqrt(51). Therefore, the answer is sqrt(51).

But maybe the problem is expecting me to compute it without using coordinates? Let me think. If I didn't have the coordinates, but just the information that sin D = 0.7, and that it's a right triangle with right angle at E, then how would I find DE?

So, in that case, let me try again. Let me denote the triangle DEF with right angle at E. So angle D is at D, angle E is the right angle, angle F is the other acute angle. Then, sin D = opposite / hypotenuse. The side opposite angle D is EF, and hypotenuse is DF. Therefore, sin D = EF / DF = 0.7.

But we need to find DE. So if we can find EF and DF, or relate them. But we need more information. However, in the diagram, EF is given as 7. Wait, in the Asymptote code, the label "7" is at the midpoint of E and F, which is the segment EF. Therefore, EF is 7. Therefore, in the problem, EF is 7. Therefore, sin D = EF / DF = 7 / DF = 0.7, so DF = 7 / 0.7 = 10. Then, using Pythagoras, DE^2 + EF^2 = DF^2 => DE^2 + 49 = 100 => DE^2 = 51 => DE = sqrt(51).

Therefore, regardless of the coordinates, if EF is 7, then DE is sqrt(51). Therefore, the answer is sqrt(51).

But let me check if there's another way to approach this. Maybe using trigonometry. For example, if we know sin D = 0.7, then we can find the other trigonometric ratios. For angle D, sin D = 0.7 = 7/10, so the opposite side is 7, hypotenuse is 10, so adjacent side is sqrt(10^2 - 7^2) = sqrt(100 - 49) = sqrt(51). Therefore, adjacent side is sqrt(51), which is DE. Therefore, DE = sqrt(51).

Yes, that's another way to see it. So regardless of the method, DE is sqrt(51). Therefore, the answer is sqrt(51).

But let me check if the problem is maybe expecting a decimal or something else? But the problem says "What is DE?" and given that in the Asymptote code, D is at (sqrt(51),7), so DE is sqrt(51). Therefore, the answer is sqrt(51).

Alternatively, maybe I need to check if there's a misunderstanding in the triangle. Let me confirm once again. The right angle is at E, so sides DE and EF are the legs, DF is the hypotenuse. So angle D is at D, so the sides relative to angle D: opposite side is EF, adjacent side is DE, hypotenuse is DF. Therefore, sin D = EF / DF = 0.7. Given EF = 7, so DF = 10. Then DE = sqrt(DF^2 - EF^2) = sqrt(100 - 49) = sqrt(51).

Yes, that's consistent. Therefore, the answer is sqrt(51).

Alternatively, if someone thought that angle D is at the right angle, but no, the right angle is at E. So angle D is one of the acute angles.

Alternatively, maybe someone could confuse the sides. For example, if they thought that DE is the hypotenuse, but no, because the hypotenuse is opposite the right angle, which is at E, so hypotenuse is DF.

Therefore, I think my reasoning is correct. Therefore, DE is sqrt(51).

**Final Answer**
\boxed{\sqrt{51}}

To determine the length of segment $ DE $ in triangle $ DEF $, we are given the following:

- Triangle $ DEF $ is a right triangle with a right angle at point $ E $.
- $ \sin D = 0.7 $
- Segment $ EF $ is labeled as 7 units long in the diagram.

---

### Step 1: Understand the Triangle Configuration

Since the right angle is at $ E $, the triangle has:

- **Legs**: $ DE $ and $ EF $
- **Hypotenuse**: $ DF $

Angle $ D $ is one of the acute angles of the triangle. In a right triangle, the sine of an angle is defined as:

$$
\sin D = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{EF}{DF}
$$

We are given that $ \sin D = 0.7 $ and $ EF = 7 $. Using this, we can solve for the hypotenuse $ DF $:

$$
\frac{7}{DF} = 0.7 \Rightarrow DF = \frac{7}{0.7} = 10
$$

---

### Step 2: Use the Pythagorean Theorem to Find $ DE $

Now that we know:

- $ EF = 7 $
- $ DF = 10 $

We can apply the Pythagorean Theorem to find $ DE $, the other leg of the triangle:

$$
DE^2 + EF^2 = DF^2
$$
$$
DE^2 + 7^2 = 10^2
$$
$$
DE^2 + 49 = 100
$$
$$
DE^2 = 51
$$
$$
DE = \sqrt{51}
$$

---

### Final Answer

$$
\boxed{\sqrt{51}}
$$

[Assistant]
Explanation:
Okay, so I need to find the length of DE in this diagram. Let me try to visualize it first. There's a right triangle DEF, right? Because there's a right angle mark at E, between D, E, and F. So, the right angle is at E. Let me confirm that from the Asymptote code. The Asymptote code defines points D, E, F. F is at (0,0), D is at (sqrt(51),7), and E is at (0,7). Then it draws D connected to E, E connected to F, and F connected back to D. The right angle mark is at D, E, F, so the right angle is at E. So triangle DEF is a right triangle with right angle at E.

Given that sin D = 0.7, and we need to find DE. Let me recall that in a right triangle, the sine of an angle is opposite over hypotenuse. So angle D is at point D, so let me think about the sides relative to angle D.

In triangle DEF, angle D is at vertex D. The sides of the triangle are DE, DF, and EF. Since the right angle is at E, the hypotenuse would be DF. Wait, because in a right triangle, the hypotenuse is opposite the right angle. So if the right angle is at E, then the hypotenuse is DF. Therefore, angle D is one of the acute angles.

So, sin D would be opposite over hypotenuse. The side opposite angle D is EF, and the hypotenuse is DF. Therefore, sin D = EF / DF = 0.7.

But wait, let me make sure. Let me label the triangle properly. Let me sketch mentally: point E is at (0,7), F is at (0,0), so EF is a vertical line segment from (0,0) to (0,7), so that's length 7. Then D is at (sqrt(51),7), so that's a point with the same y-coordinate as E, so DE is a horizontal line from E (0,7) to D (sqrt(51),7), so DE is horizontal. Then DF connects D (sqrt(51),7) to F (0,0). So triangle DEF has right angle at E, with legs EF (vertical) and DE (horizontal), and hypotenuse DF.

Therefore, angle D is at point D, so looking at angle D, the sides adjacent and opposite. Let me think again. For angle D, the sides are: the side opposite to angle D is EF, which is 7 units long. The hypotenuse is DF. The side adjacent to angle D is DE.

Therefore, sin D = opposite / hypotenuse = EF / DF = 7 / DF = 0.7. Therefore, 7 / DF = 0.7, so DF = 7 / 0.7 = 10. Therefore, DF is 10.

But wait, the problem is asking for DE. So maybe I need to find DE? But if I can find DF, which is 10, and I know EF is 7, then since DEF is a right triangle, by Pythagoras, DE^2 + EF^2 = DF^2. Therefore, DE^2 + 7^2 = 10^2. Therefore, DE^2 + 49 = 100. Therefore, DE^2 = 51. Therefore, DE = sqrt(51).

Wait, but the coordinates of D are given as (sqrt(51),7). Since E is at (0,7), then DE is the horizontal distance between (sqrt(51),7) and (0,7), which is sqrt(51) - 0 = sqrt(51). So DE is sqrt(51). Therefore, the answer is sqrt(51).

But let me check if this is consistent with the given sin D = 0.7. Let me verify. If DE is sqrt(51), EF is 7, DF is 10. Then sin D is EF / DF = 7 / 10 = 0.7, which matches the given value. So that seems correct.

But wait, the problem says "Suppose sin D = 0.7 in the diagram below." But in the Asymptote code, they have coordinates for D, E, F. So maybe they are giving coordinates, but the problem is just asking for DE? But according to the coordinates, DE is sqrt(51) - 0 = sqrt(51). But maybe the problem is not using the coordinates? Wait, the Asymptote code is just for the diagram, but the problem says "Suppose sin D = 0.7 in the diagram below." So maybe the diagram is just a generic diagram, and the coordinates are just for drawing? Wait, but the Asymptote code defines specific coordinates. Let me check again.

Wait, the Asymptote code defines F at (0,0), D at (sqrt(51),7), E at (0,7). So in that diagram, DE is sqrt(51), EF is 7, DF is sqrt( (sqrt(51))^2 + 7^2 ) = sqrt(51 + 49) = sqrt(100) = 10. Therefore, sin D is EF / DF = 7 / 10 = 0.7, which matches the given value. Therefore, in this diagram, DE is sqrt(51). Therefore, the answer is sqrt(51).

But maybe the problem is expecting me to compute it without using coordinates? Let me think. If I didn't have the coordinates, but just the information that sin D = 0.7, and that it's a right triangle with right angle at E, then how would I find DE?

So, in that case, let me try again. Let me denote the triangle DEF with right angle at E. So angle D is at D, angle E is the right angle, angle F is the other acute angle. Then, sin D = opposite / hypotenuse. The side opposite angle D is EF, and hypotenuse is DF. Therefore, sin D = EF / DF = 0.7.

But we need to find DE. So if we can find EF and DF, or relate them. But we need more information. However, in the diagram, EF is given as 7. Wait, in the Asymptote code, the label "7" is at the midpoint of E and F, which is the segment EF. Therefore, EF is 7. Therefore, in the problem, EF is 7. Therefore, sin D = EF / DF = 7 / DF = 0.7, so DF = 7 / 0.7 = 10. Then, using Pythagoras, DE^2 + EF^2 = DF^2 => DE^2 + 49 = 100 => DE^2 = 51 => DE = sqrt(51).

Therefore, regardless of the coordinates, if EF is 7, then DE is sqrt(51). Therefore, the answer is sqrt(51).

But let me check if there's another way to approach this. Maybe using trigonometry. For example, if we know sin D = 0.7, then we can find the other trigonometric ratios. For angle D, sin D = 0.7 = 7/10, so the opposite side is 7, hypotenuse is 10, so adjacent side is sqrt(10^2 - 7^2) = sqrt(100 - 49) = sqrt(51). Therefore, adjacent side is sqrt(51), which is DE. Therefore, DE = sqrt(51).

Yes, that's another way to see it. So regardless of the method, DE is sqrt(51). Therefore, the answer is sqrt(51).

But let me check if the problem is maybe expecting a decimal or something else? But the problem says "What is DE?" and given that in the Asymptote code, D is at (sqrt(51),7), so DE is sqrt(51). Therefore, the answer is sqrt(51).

Alternatively, maybe I need to check if there's a misunderstanding in the triangle. Let me confirm once again. The right angle is at E, so sides DE and EF are the legs, DF is the hypotenuse. So angle D is at D, so the sides relative to angle D: opposite side is EF, adjacent side is DE, hypotenuse is DF. Therefore, sin D = EF / DF = 0.7. Given EF = 7, so DF = 10. Then DE = sqrt(DF^2 - EF^2) = sqrt(100 - 49) = sqrt(51).

Yes, that's consistent. Therefore, the answer is sqrt(51).

Alternatively, if someone thought that angle D is at the right angle, but no, the right angle is at E. So angle D is one of the acute angles.

Alternatively, maybe someone could confuse the sides. For example, if they thought that DE is the hypotenuse, but no, because the hypotenuse is opposite the right angle, which is at E, so hypotenuse is DF.

Therefore, I think my reasoning is correct. Therefore, DE is sqrt(51).

**Final Answer**
\boxed{\sqrt{51}}

To determine the length of segment $ DE $ in triangle $ DEF $, we are given the following:

- Triangle $ DEF $ is a right triangle with a right angle at point $ E $.
- $ \sin D = 0.7 $
- Segment $ EF $ is labeled as 7 units long in the diagram.

---

### Step 1: Understand the Triangle Configuration

Since the right angle is at $ E $, the triangle has:

- **Legs**: $ DE $ and $ EF $
- **Hypotenuse**: $ DF $

Angle $ D $ is one of the acute angles of the triangle. In a right triangle, the sine of an angle is defined as:

$$
\sin D = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{EF}{DF}
$$

We are given that $ \sin D = 0.7 $ and $ EF = 7 $. Using this, we can solve for the hypotenuse $ DF $:

$$
\frac{7}{DF} = 0.7 \Rightarrow DF = \frac{7}{0.7} = 10
$$

---

### Step 2: Use the Pythagorean Theorem to Find $ DE $

Now that we know:

- $ EF = 7 $
- $ DF = 10 $

We can apply the Pythagorean Theorem to find $ DE $, the other leg of the triangle:

$$
DE^2 + EF^2 = DF^2
$$
$$
DE^2 + 7^2 = 10^2
$$
$$
DE^2 + 49 = 100
$$
$$
DE^2 = 51
$$
$$
DE = \sqrt{51}
$$

---

### Final Answer

$$
\boxed{\sqrt{51}}
$$

Solution:
The triangle is a right triangle, so $\sin D = \frac{EF}{DF}$. Then we have that $\sin D = 0.7 = \frac{7}{DF}$, so $DF = 10$.

Using the Pythagorean Theorem, we find that the length of $DE$ is $\sqrt{DF^2 - EF^2},$ or $\sqrt{100 - 49} = \boxed{\sqrt{51}}$.


Explanation:
Okay, so I need to compute this series: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Let me think about how to approach this.

First, let me write down the series to visualize it better. It starts with 1, then subtracts 2, adds 3, subtracts 4, and so on, alternating between adding and subtracting each subsequent integer. The last term is -100. So, the pattern is +1 -2 +3 -4 +5 -6 ... +99 -100.

Hmm, since the series alternates between positive and negative terms, maybe I can group them in pairs? Let me try that. If I group the first two terms, then the next two, and so on, each pair would be something like (1 - 2), (3 - 4), (5 - 6), ..., (99 - 100).

Let me check how many such pairs there are. The series goes up to 100, so if I pair them as (1-2), (3-4), ..., (99-100), each pair consists of two numbers. Since there are 100 terms in total, that would make 50 pairs. Let me confirm: 100 divided by 2 is 50. Yep, that seems right.

Now, let me compute the value of each pair. Take the first pair: 1 - 2 = -1. The second pair: 3 - 4 = -1. Similarly, the third pair: 5 - 6 = -1. Wait a minute, every pair seems to equal -1. So if each of the 50 pairs is -1, then the total sum would be 50 multiplied by -1, which is -50.

But let me make sure I didn't make a mistake here. Let me check with a smaller number of terms. Let's say if the series was 1 - 2 + 3 - 4. That would be (1 - 2) + (3 - 4) = (-1) + (-1) = -2. Which is indeed -2, and since there are 4 terms, 4/2 = 2 pairs, so 2*(-1) = -2. That works.

Another example: 1 - 2 + 3 - 4 + 5 - 6. That would be three pairs: (1-2) + (3-4) + (5-6) = -1 + -1 + -1 = -3. Which is 3*(-1) = -3. So that seems consistent.

Therefore, applying this to the original problem, since there are 100 terms, 50 pairs, each summing to -1, the total sum should be -50.

But let me think if there's another way to approach this problem, just to verify. Maybe using formulas for arithmetic series or something else.

The original series can be considered as the sum from n=1 to n=100 of (-1)^{n+1} * n. Wait, let me check: when n=1, (-1)^{1+1} *1 = (-1)^2 *1 = 1*1=1. Then n=2: (-1)^{3}*2 = -1*2=-2. Yep, that matches the series.

Alternatively, maybe I can separate the series into the sum of the positive terms and the sum of the negative terms. Let's try that.

The positive terms are the odd-numbered terms: 1, 3, 5, ..., 99. The negative terms are the even-numbered terms: -2, -4, -6, ..., -100.

So the total sum is (1 + 3 + 5 + ... + 99) + (-2 -4 -6 - ... -100).

Let me compute each part separately.

First, the sum of the positive terms: 1 + 3 + 5 + ... + 99. This is the sum of the first 50 odd numbers. Wait, because the odd numbers go from 1 to 99. How many terms is that? Well, since 99 is the 50th odd number (since 2*50 -1 = 99). So there are 50 terms.

There's a formula for the sum of the first n odd numbers: n^2. So, sum = 50^2 = 2500.

Wait, let me confirm that formula. The sum of the first n odd numbers is indeed n². For example, 1 = 1², 1+3=4=2², 1+3+5=9=3², etc. So yes, that formula holds. Therefore, the sum of the positive terms is 50² = 2500.

Now, the sum of the negative terms: -2 -4 -6 - ... -100. Let me factor out the negative sign: - (2 + 4 + 6 + ... + 100).

Now, the sum inside the parentheses is the sum of the first 50 even numbers. Because 2, 4, 6, ..., 100. How many terms? Well, 100 is the 50th even number (since 2*50=100).

The formula for the sum of the first n even numbers is n(n + 1). Wait, let me recall. The sum of the first n even numbers is n(n + 1). Let me check with small numbers. For n=1: 2 = 1*(1+1)=2. Correct. For n=2: 2 + 4 = 6, and 2*(2+1)=6. Correct. For n=3: 2+4+6=12, 3*(3+1)=12. Correct. So yes, the formula is n(n + 1).

Therefore, the sum of the first 50 even numbers is 50*51 = 2550. Therefore, the negative terms sum to -2550.

Therefore, the total sum is 2500 (positive terms) + (-2550) (negative terms) = 2500 - 2550 = -50.

Okay, so that matches the previous result. So that's reassuring.

Alternatively, maybe I can use the formula for the sum of an alternating series. But I think the two methods I used are sufficient.

Wait, let me think again. Another way: Let me consider the general formula for the sum of such an alternating series.

Suppose we have S = 1 - 2 + 3 - 4 + 5 - ... + (-1)^{n+1} n.

In our case, n=100.

There is a formula for this sum. Let me recall.

If n is even, then the sum S can be written as -n/2. Wait, let me check with n=2: 1 - 2 = -1, which is -2/2 = -1. Correct. For n=4: 1 -2 +3 -4 = -2, which is -4/2 = -2. Correct. For n=6: 1 -2 +3 -4 +5 -6 = -3, which is -6/2 = -3. So yes, if n is even, the sum is -n/2.

Therefore, in our case, n=100, which is even, so the sum is -100/2 = -50.

So that's another way to see it.

Therefore, regardless of the method, the answer is -50.

But let me just make sure that this formula is correct. Let me try to derive it.

Suppose S = 1 - 2 + 3 - 4 + ... + (n-1) - n, where n is even.

Group the terms as (1 - 2) + (3 - 4) + ... + ((n-1) - n). There are n/2 groups, each of which is -1. Therefore, total sum is (n/2)*(-1) = -n/2.

Yes, that's exactly what we did earlier. So that's the derivation.

Therefore, for even n, the sum is -n/2.

Therefore, for n=100, it's -100/2 = -50.

So all methods lead to the same answer.

Alternatively, if I didn't remember the formula, I could think of it as follows:

Let me write S = 1 - 2 + 3 - 4 + ... + 99 - 100.

Let me consider adding S to itself but shifting the terms. Wait, that might be more complicated.

Alternatively, think of S as the sum from k=1 to 100 of (-1)^{k+1} k.

But maybe another approach. Let me think.

Alternatively, consider that the series is similar to the alternating harmonic series, but with integers.

But perhaps not necessary here.

Alternatively, think of the sum as the difference between the sum of odd numbers and the sum of even numbers. Which is exactly what I did earlier.

So, sum of odd numbers up to 99 is 50² = 2500. Sum of even numbers up to 100 is 50*51 = 2550. Then the difference is 2500 - 2550 = -50.

So, again, same answer.

Therefore, I think I can be confident that the answer is -50.

But just to make sure, let me compute a few more terms manually.

For example, take n=6: 1 -2 +3 -4 +5 -6 = (1-2) + (3-4) + (5-6) = (-1) + (-1) + (-1) = -3. Which is -6/2 = -3. Correct.

Another example, n=10: 1-2+3-4+5-6+7-8+9-10. Grouping: (1-2)+(3-4)+(5-6)+(7-8)+(9-10) = (-1)*5 = -5. Which is -10/2 = -5. Correct.

So, seems like the formula holds.

Therefore, applying it to n=100, answer is -50.

Alternatively, if I didn't think of grouping or formulas, maybe I can think recursively.

Suppose I know the sum up to some number. For example, if I have S(n) = 1 - 2 + 3 - ... + (-1)^{n+1} n.

If n is even, then S(n) = S(n-2) + (n-1) - n.

But maybe that complicates.

Alternatively, think of S(n) = S(n-1) + (-1)^{n+1} n.

But again, maybe not helpful here.

Alternatively, think of the series as two separate series:

Sum_{k=1}^{50} (2k - 1) - Sum_{k=1}^{50} 2k.

Which is exactly what I did earlier.

So, Sum_{k=1}^{50} (2k - 1) is the sum of odd numbers, which is 50² = 2500.

Sum_{k=1}^{50} 2k is 2 * Sum_{k=1}^{50} k = 2*(50*51)/2 = 50*51 = 2550.

Therefore, 2500 - 2550 = -50.

Same result.

Therefore, regardless of the method, the answer is -50.

I think that's thorough enough. I can't see any mistakes in my reasoning. All different approaches lead to the same conclusion.

Therefore, the final answer is -50.

**Final Answer**
\boxed{-50}

To compute the sum of the alternating series:

$$
1 - 2 + 3 - 4 + 5 - 6 + \dots + 99 - 100
$$

we observe that the series alternates between positive and negative terms, starting with a positive term. The pattern is:

$$
+1 - 2 + 3 - 4 + 5 - 6 + \dots + 99 - 100
$$

---

### Step 1: Group the Terms

We can group the terms in pairs:

$$
(1 - 2) + (3 - 4) + (5 - 6) + \dots + (99 - 100)
$$

Each pair simplifies to:

$$
1 - 2 = -1,\quad 3 - 4 = -1,\quad 5 - 6 = -1,\quad \dots,\quad 99 - 100 = -1
$$

There are 100 terms in total, so there are:

$$
\frac{100}{2} = 50 \text{ pairs}
$$

Each pair contributes $-1$, so the total sum is:

$$
50 \times (-1) = -50
$$

---

### Step 2: Alternative Verification Using Sum of Odd and Even Numbers

We can also separate the series into the sum of **positive terms** (odd numbers) and the sum of **negative terms** (even numbers):

- **Positive terms**: $1 + 3 + 5 + \dots + 99$
- **Negative terms**: $-2 - 4 - 6 - \dots - 100$

#### Sum of Positive Terms (Odd Numbers)

The first 50 odd numbers are:

$$
1, 3, 5, \dots, 99
$$

The sum of the first $n$ odd numbers is given by the formula:

$$
\text{Sum} = n^2
$$

So, the sum of the first 50 odd numbers is:

$$
50^2 = 2500
$$

#### Sum of Negative Terms (Even Numbers)

The first 50 even numbers are:

$$
2, 4, 6, \dots, 100
$$

The sum of the first $n$ even numbers is:

$$
\text{Sum} = n(n + 1)
$$

So, the sum of the first 50 even numbers is:

$$
50 \times 51 = 2550
$$

Since these are negative, the total contribution is:

$$
-2550
$$

#### Total Sum

Now, adding the two parts:

$$
2500 - 2550 = -50
$$

---

### Step 3: General Formula for Alternating Series

This series is a special case of the alternating sum:

$$
S = \sum_{k=1}^{n} (-1)^{k+1} k
$$

For even $n$, the sum simplifies to:

$$
S = -\frac{n}{2}
$$

In our case, $n = 100$, so:

$$
S = -\frac{100}{2} = -50
$$

---

### Final Answer

$$
\boxed{-50}
$$

[Assistant]
Explanation:
Okay, so I need to compute this series: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Let me think about how to approach this.

First, let me write down the series to visualize it better. It starts with 1, then subtracts 2, adds 3, subtracts 4, and so on, alternating between adding and subtracting each subsequent integer. The last term is -100. So, the pattern is +1 -2 +3 -4 +5 -6 ... +99 -100.

Hmm, since the series alternates between positive and negative terms, maybe I can group them in pairs? Let me try that. If I group the first two terms, then the next two, and so on, each pair would be something like (1 - 2), (3 - 4), (5 - 6), ..., (99 - 100).

Let me check how many such pairs there are. The series goes up to 100, so if I pair them as (1-2), (3-4), ..., (99-100), each pair consists of two numbers. Since there are 100 terms in total, that would make 50 pairs. Let me confirm: 100 divided by 2 is 50. Yep, that seems right.

Now, let me compute the value of each pair. Take the first pair: 1 - 2 = -1. The second pair: 3 - 4 = -1. Similarly, the third pair: 5 - 6 = -1. Wait a minute, every pair seems to equal -1. So if each of the 50 pairs is -1, then the total sum would be 50 multiplied by -1, which is -50.

But let me make sure I didn't make a mistake here. Let me check with a smaller number of terms. Let's say if the series was 1 - 2 + 3 - 4. That would be (1 - 2) + (3 - 4) = (-1) + (-1) = -2. Which is indeed -2, and since there are 4 terms, 4/2 = 2 pairs, so 2*(-1) = -2. That works.

Another example: 1 - 2 + 3 - 4 + 5 - 6. That would be three pairs: (1-2) + (3-4) + (5-6) = -1 + -1 + -1 = -3. Which is 3*(-1) = -3. So that seems consistent.

Therefore, applying this to the original problem, since there are 100 terms, 50 pairs, each summing to -1, the total sum should be -50.

But let me think if there's another way to approach this problem, just to verify. Maybe using formulas for arithmetic series or something else.

The original series can be considered as the sum from n=1 to n=100 of (-1)^{n+1} * n. Wait, let me check: when n=1, (-1)^{1+1} *1 = (-1)^2 *1 = 1*1=1. Then n=2: (-1)^{3}*2 = -1*2=-2. Yep, that matches the series.

Alternatively, maybe I can separate the series into the sum of the positive terms and the sum of the negative terms. Let's try that.

The positive terms are the odd-numbered terms: 1, 3, 5, ..., 99. The negative terms are the even-numbered terms: -2, -4, -6, ..., -100.

So the total sum is (1 + 3 + 5 + ... + 99) + (-2 -4 -6 - ... -100).

Let me compute each part separately.

First, the sum of the positive terms: 1 + 3 + 5 + ... + 99. This is the sum of the first 50 odd numbers. Wait, because the odd numbers go from 1 to 99. How many terms is that? Well, since 99 is the 50th odd number (since 2*50 -1 = 99). So there are 50 terms.

There's a formula for the sum of the first n odd numbers: n^2. So, sum = 50^2 = 2500.

Wait, let me confirm that formula. The sum of the first n odd numbers is indeed n². For example, 1 = 1², 1+3=4=2², 1+3+5=9=3², etc. So yes, that formula holds. Therefore, the sum of the positive terms is 50² = 2500.

Now, the sum of the negative terms: -2 -4 -6 - ... -100. Let me factor out the negative sign: - (2 + 4 + 6 + ... + 100).

Now, the sum inside the parentheses is the sum of the first 50 even numbers. Because 2, 4, 6, ..., 100. How many terms? Well, 100 is the 50th even number (since 2*50=100).

The formula for the sum of the first n even numbers is n(n + 1). Wait, let me recall. The sum of the first n even numbers is n(n + 1). Let me check with small numbers. For n=1: 2 = 1*(1+1)=2. Correct. For n=2: 2 + 4 = 6, and 2*(2+1)=6. Correct. For n=3: 2+4+6=12, 3*(3+1)=12. Correct. So yes, the formula is n(n + 1).

Therefore, the sum of the first 50 even numbers is 50*51 = 2550. Therefore, the negative terms sum to -2550.

Therefore, the total sum is 2500 (positive terms) + (-2550) (negative terms) = 2500 - 2550 = -50.

Okay, so that matches the previous result. So that's reassuring.

Alternatively, maybe I can use the formula for the sum of an alternating series. But I think the two methods I used are sufficient.

Wait, let me think again. Another way: Let me consider the general formula for the sum of such an alternating series.

Suppose we have S = 1 - 2 + 3 - 4 + 5 - ... + (-1)^{n+1} n.

In our case, n=100.

There is a formula for this sum. Let me recall.

If n is even, then the sum S can be written as -n/2. Wait, let me check with n=2: 1 - 2 = -1, which is -2/2 = -1. Correct. For n=4: 1 -2 +3 -4 = -2, which is -4/2 = -2. Correct. For n=6: 1 -2 +3 -4 +5 -6 = -3, which is -6/2 = -3. So yes, if n is even, the sum is -n/2.

Therefore, in our case, n=100, which is even, so the sum is -100/2 = -50.

So that's another way to see it.

Therefore, regardless of the method, the answer is -50.

But let me just make sure that this formula is correct. Let me try to derive it.

Suppose S = 1 - 2 + 3 - 4 + ... + (n-1) - n, where n is even.

Group the terms as (1 - 2) + (3 - 4) + ... + ((n-1) - n). There are n/2 groups, each of which is -1. Therefore, total sum is (n/2)*(-1) = -n/2.

Yes, that's exactly what we did earlier. So that's the derivation.

Therefore, for even n, the sum is -n/2.

Therefore, for n=100, it's -100/2 = -50.

So all methods lead to the same answer.

Alternatively, if I didn't remember the formula, I could think of it as follows:

Let me write S = 1 - 2 + 3 - 4 + ... + 99 - 100.

Let me consider adding S to itself but shifting the terms. Wait, that might be more complicated.

Alternatively, think of S as the sum from k=1 to 100 of (-1)^{k+1} k.

But maybe another approach. Let me think.

Alternatively, consider that the series is similar to the alternating harmonic series, but with integers.

But perhaps not necessary here.

Alternatively, think of the sum as the difference between the sum of odd numbers and the sum of even numbers. Which is exactly what I did earlier.

So, sum of odd numbers up to 99 is 50² = 2500. Sum of even numbers up to 100 is 50*51 = 2550. Then the difference is 2500 - 2550 = -50.

So, again, same answer.

Therefore, I think I can be confident that the answer is -50.

But just to make sure, let me compute a few more terms manually.

For example, take n=6: 1 -2 +3 -4 +5 -6 = (1-2) + (3-4) + (5-6) = (-1) + (-1) + (-1) = -3. Which is -6/2 = -3. Correct.

Another example, n=10: 1-2+3-4+5-6+7-8+9-10. Grouping: (1-2)+(3-4)+(5-6)+(7-8)+(9-10) = (-1)*5 = -5. Which is -10/2 = -5. Correct.

So, seems like the formula holds.

Therefore, applying it to n=100, answer is -50.

Alternatively, if I didn't think of grouping or formulas, maybe I can think recursively.

Suppose I know the sum up to some number. For example, if I have S(n) = 1 - 2 + 3 - ... + (-1)^{n+1} n.

If n is even, then S(n) = S(n-2) + (n-1) - n.

But maybe that complicates.

Alternatively, think of S(n) = S(n-1) + (-1)^{n+1} n.

But again, maybe not helpful here.

Alternatively, think of the series as two separate series:

Sum_{k=1}^{50} (2k - 1) - Sum_{k=1}^{50} 2k.

Which is exactly what I did earlier.

So, Sum_{k=1}^{50} (2k - 1) is the sum of odd numbers, which is 50² = 2500.

Sum_{k=1}^{50} 2k is 2 * Sum_{k=1}^{50} k = 2*(50*51)/2 = 50*51 = 2550.

Therefore, 2500 - 2550 = -50.

Same result.

Therefore, regardless of the method, the answer is -50.

I think that's thorough enough. I can't see any mistakes in my reasoning. All different approaches lead to the same conclusion.

Therefore, the final answer is -50.

**Final Answer**
\boxed{-50}

To compute the sum of the alternating series:

$$
1 - 2 + 3 - 4 + 5 - 6 + \dots + 99 - 100
$$

we observe that the series alternates between positive and negative terms, starting with a positive term. The pattern is:

$$
+1 - 2 + 3 - 4 + 5 - 6 + \dots + 99 - 100
$$

---

### Step 1: Group the Terms

We can group the terms in pairs:

$$
(1 - 2) + (3 - 4) + (5 - 6) + \dots + (99 - 100)
$$

Each pair simplifies to:

$$
1 - 2 = -1,\quad 3 - 4 = -1,\quad 5 - 6 = -1,\quad \dots,\quad 99 - 100 = -1
$$

There are 100 terms in total, so there are:

$$
\frac{100}{2} = 50 \text{ pairs}
$$

Each pair contributes $-1$, so the total sum is:

$$
50 \times (-1) = -50
$$

---

### Step 2: Alternative Verification Using Sum of Odd and Even Numbers

We can also separate the series into the sum of **positive terms** (odd numbers) and the sum of **negative terms** (even numbers):

- **Positive terms**: $1 + 3 + 5 + \dots + 99$
- **Negative terms**: $-2 - 4 - 6 - \dots - 100$

#### Sum of Positive Terms (Odd Numbers)

The first 50 odd numbers are:

$$
1, 3, 5, \dots, 99
$$

The sum of the first $n$ odd numbers is given by the formula:

$$
\text{Sum} = n^2
$$

So, the sum of the first 50 odd numbers is:

$$
50^2 = 2500
$$

#### Sum of Negative Terms (Even Numbers)

The first 50 even numbers are:

$$
2, 4, 6, \dots, 100
$$

The sum of the first $n$ even numbers is:

$$
\text{Sum} = n(n + 1)
$$

So, the sum of the first 50 even numbers is:

$$
50 \times 51 = 2550
$$

Since these are negative, the total contribution is:

$$
-2550
$$

#### Total Sum

Now, adding the two parts:

$$
2500 - 2550 = -50
$$

---

### Step 3: General Formula for Alternating Series

This series is a special case of the alternating sum:

$$
S = \sum_{k=1}^{n} (-1)^{k+1} k
$$

For even $n$, the sum simplifies to:

$$
S = -\frac{n}{2}
$$

In our case, $n = 100$, so:

$$
S = -\frac{100}{2} = -50
$$

---

### Final Answer

$$
\boxed{-50}
$$

Solution: $(1-2)+(3-4)+ \dots +(97-98)+(99-100) = 50(-1) = \boxed{-50}.$