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 nth odd number is 2n - 1. So if the last term is 99, then 2n - 1 = 99 => 2n = 100 => n = 50. So there are 50 terms here.

I remember that the sum of the first m odd numbers is m². Let me verify that. For example, sum of first 1 odd number: 1 = 1². First 2: 1 + 3 = 4 = 2². First 3: 1 + 3 + 5 = 9 = 3². Yep, that formula works. So the sum of the first 50 odd numbers 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).

The sum inside the parentheses is the sum of the first 50 even numbers. Because 2, 4, ..., 100. The nth even number is 2n. So the sum from n=1 to 50 of 2n.

Which is 2*(1 + 2 + 3 + ... + 50). The sum of the first m integers is m(m + 1)/2. So here, m = 50. Therefore, the sum is 2*(50*51)/2. The 2 and the denominator 2 cancel out, so it's 50*51 = 2550.

Therefore, the negative terms sum to -2550.

So the total sum is the sum of positive terms (2500) plus the sum of negative terms (-2550).

So 2500 - 2550 = -50.

Same result as before! So that confirms that the answer is indeed -50.

Alternatively, maybe I can use another formula. Let me recall that for an alternating series like this, there might be a general formula.

But since I already have two methods giving me -50, maybe I can be confident. However, let me try another approach just to be thorough.

Another way: Let me consider the series S = 1 - 2 + 3 - 4 + 5 - ... + 99 - 100.

If I write this as S = Σ_{k=1}^{100} (-1)^{k+1} * k.

Alternatively, I can think of this as S = Σ_{k=1}^{100} (-1)^{k+1} * k.

But maybe I can relate this to the formula for the sum of an alternating series. However, I don't remember a specific formula for this, but perhaps I can derive it.

Alternatively, consider that for an alternating series with terms alternating sign, if the number of terms is even, then we can pair them as (1 - 2) + (3 - 4) + ... + (99 - 100) as before, which gives -1 * (number of pairs). Since there are 100 terms, 50 pairs, so -50.

Alternatively, if I think of the general formula for the sum of such a series. Let me think.

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

If n is even, say n = 2m, then S = Σ_{k=1}^{2m} (-1)^{k+1} k.

Which can be written as Σ_{j=1}^{m} [ (2j - 1) - 2j ] = Σ_{j=1}^{m} (-1) = -m.

Since n = 2m, m = n/2. Therefore, S = -n/2.

In our case, n = 100, so S = -100/2 = -50.

Exactly! So that's another way to see it. Therefore, the answer is -50.

So all methods lead to the same conclusion.

But just to make sure, let me compute the sum step by step for a slightly larger number and see if the pattern holds.

For example, let's take n = 4: 1 - 2 + 3 - 4 = (1 - 2) + (3 - 4) = -1 + -1 = -2. According to the formula, -n/2 = -4/2 = -2. Correct.

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

So the formula seems to hold. Therefore, for n = 100, it's -100/2 = -50.

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

But just to make sure, let me try to compute the sum using another method. Let me consider the general formula for the sum of an alternating series.

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

This can be written as S = Σ_{k=1}^n (-1)^{k+1} k.

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

If we consider the sum S = Σ_{k=1}^n (-1)^{k+1} k.

Let me consider two cases: when n is even and when n is odd.

Case 1: n is even. Let n = 2m. Then S = Σ_{k=1}^{2m} (-1)^{k+1} k = Σ_{j=1}^m [ (2j - 1) - 2j ] = Σ_{j=1}^m (-1) = -m.

Which is -n/2.

Case 2: n is odd. Let n = 2m + 1. Then S = Σ_{k=1}^{2m+1} (-1)^{k+1} k = Σ_{j=1}^m [ (2j - 1) - 2j ] + (2m + 1) = Σ_{j=1}^m (-1) + (2m + 1) = -m + 2m + 1 = m + 1.

Which is (n + 1)/2.

So in our problem, n = 100, which is even, so S = -n/2 = -50.

Therefore, this formula confirms again that the answer is -50.

Alternatively, if I didn't remember the formula, I could derive it.

Let me try to derive it. Let me consider S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me write S as follows:

If n is even, then S = (1 - 2) + (3 - 4) + ... + (n-1 - n).

Each pair is -1, and there are n/2 pairs. Therefore, S = -n/2.

If n is odd, then S = (1 - 2) + (3 - 4) + ... + (n-2 - (n-1)) + n.

Each of the first (n-1)/2 pairs is -1, and then the last term is n. Therefore, S = - (n - 1)/2 + n = (-n + 1)/2 + n = (n + 1)/2.

So again, same result.

Therefore, since our problem has n = 100, which is even, the answer is -100/2 = -50.

Another way to think about it is using recursion or recurrence relations. Let me try that.

Let me denote S(n) as the sum up to n terms.

If n = 1, S(1) = 1.

If n = 2, S(2) = 1 - 2 = -1.

If n = 3, S(3) = 1 - 2 + 3 = 2.

If n = 4, S(4) = 1 - 2 + 3 - 4 = -2.

If n = 5, S(5) = 1 - 2 + 3 - 4 + 5 = 3.

So, it seems that for even n, S(n) = -n/2, and for odd n, S(n) = (n + 1)/2.

So for n = 100, which is even, S(100) = -100/2 = -50.

Therefore, this pattern also gives the same answer.

Alternatively, maybe I can use generating functions or other advanced methods, but that might be overkill.

Alternatively, think of the series as the difference between the sum of odd numbers and the sum of even numbers.

As I did earlier, the sum of odd numbers from 1 to 99 is 50² = 2500. The sum of even numbers from 2 to 100 is 2 + 4 + ... + 100. Which is 2*(1 + 2 + ... + 50) = 2*(50*51)/2 = 50*51 = 2550. Therefore, the total sum is 2500 - 2550 = -50.

Same result again.

So, all methods converge to -50.

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

But just to make sure, let me compute the sum manually for a small n and see.

Take n = 2: 1 - 2 = -1. According to formula, -2/2 = -1. Correct.

n = 4: 1 - 2 + 3 - 4 = -2. Formula: -4/2 = -2. Correct.

n = 6: 1 - 2 + 3 - 4 + 5 - 6 = -3. Formula: -6/2 = -3. Correct.

n = 100: -100/2 = -50.

Therefore, the answer is definitely -50.

Alternatively, if I think about the series as telescoping or something else? Not sure. But since all methods give the same answer, I think that's solid.

Another way: Let me consider the general term.

Let me write the series as S = Σ_{k=1}^{100} (-1)^{k+1} k.

Let me consider the sum S = Σ_{k=1}^{n} (-1)^{k+1} k.

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

If I consider the sum S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

This is an alternating series.

Let me consider the formula for the sum.

Let me recall that for an alternating series, the sum can be expressed as:

If n is even: S = -n/2

If n is odd: S = (n + 1)/2

Which is exactly what we derived earlier.

Therefore, since n = 100 is even, S = -100/2 = -50.

Therefore, the answer is -50.

Alternatively, if I didn't remember the formula, I could derive it using the method of differences.

Let me consider S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me consider multiplying S by -1:

- S = -1 + 2 - 3 + 4 - ... + (-1)^{n} n.

Now, if I add S and -S:

S + (-S) = 0 = [1 - 2 + 3 - 4 + ... + (-1)^{n+1} n] + [-1 + 2 - 3 + 4 - ... + (-1)^{n} n]

Let me compute term by term.

First term: 1 + (-1) = 0

Second term: -2 + 2 = 0

Third term: 3 + (-3) = 0

Fourth term: -4 + 4 = 0

...

Continuing this way, each pair cancels out.

But wait, how many terms are there? If n is even, then the last term in S is (-1)^{n+1} n = (-1)^{even + 1} n = (-1)^{odd} n = -n.

In -S, the last term is (-1)^{n} n = (-1)^{even} n = +n.

Therefore, adding S and -S:

All terms cancel except for the last term of S and the last term of -S? Wait, no. Wait, let me check again.

Wait, if n is even, then S has terms up to (-1)^{n+1} n = -n.

And -S has terms up to (-1)^{n} n = +n.

But when adding S and -S, each term cancels except for the last term? Wait, no. Let me think again.

Wait, if n is even, then the number of terms is even. Let me take n = 4 as an example.

S = 1 - 2 + 3 - 4

- S = -1 + 2 - 3 + 4

Adding them: (1 - 1) + (-2 + 2) + (3 - 3) + (-4 + 4) = 0 + 0 + 0 + 0 = 0.

So all terms cancel. Therefore, 0 = S + (-S) = 0. Which is trivial.

But how does that help me?

Alternatively, maybe I need to consider another approach.

Alternatively, let me consider S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me consider S = Σ_{k=1}^n (-1)^{k+1} k.

Let me consider the sum S + S.

Wait, maybe not. Alternatively, let me consider S = Σ_{k=1}^n (-1)^{k+1} k.

Let me write S as:

S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me consider S = Σ_{k=1}^n (-1)^{k+1} k.

Let me consider the difference between S and the sum of all numbers from 1 to n.

Wait, not sure.

Alternatively, think of S as the difference between the sum of odd terms and the sum of even terms.

Which is exactly what I did earlier.

Sum of odd terms: 1 + 3 + 5 + ... + 99 = 2500

Sum of even terms: 2 + 4 + 6 + ... + 100 = 2550

Therefore, S = 2500 - 2550 = -50.

Therefore, same result.

Alternatively, if I didn't remember the formula for the sum of odd or even numbers, I could compute them using arithmetic series formulas.

Sum of odd numbers: first term a1 = 1, last term an = 99, number of terms m = 50.

Sum = m*(a1 + an)/2 = 50*(1 + 99)/2 = 50*100/2 = 50*50 = 2500.

Sum of even numbers: first term a1 = 2, last term an = 100, number of terms m = 50.

Sum = m*(a1 + an)/2 = 50*(2 + 100)/2 = 50*102/2 = 50*51 = 2550.

Therefore, difference is 2500 - 2550 = -50.

Same result.

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

Therefore, I can be confident that the answer is \boxed{-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. This pattern continues up to the 100th term, which is negative.

---

### Step 1: Group the Terms in Pairs

We can group the terms in pairs to simplify the computation:

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

Each pair has the form $ (2k - 1) - 2k = -1 $, and since there are 100 terms, there are $ \frac{100}{2} = 50 $ such pairs.

So the total sum becomes:

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

---

### Step 2: Confirm Using Sum of Odd and Even Numbers

We can also compute the sum by separating the positive and negative terms:

- **Positive terms (odd numbers):** $ 1 + 3 + 5 + \dots + 99 $
- **Negative terms (even numbers):** $ -2 - 4 - 6 - \dots - 100 $

#### Sum of the first 50 odd numbers:

The sum of the first $ n $ odd numbers is $ n^2 $. Here, $ n = 50 $, so:

$$
1 + 3 + 5 + \dots + 99 = 50^2 = 2500
$$

#### Sum of the first 50 even numbers:

The sum of the first $ n $ even numbers is $ n(n + 1) $. Here, $ n = 50 $, so:

$$
2 + 4 + 6 + \dots + 100 = 2(1 + 2 + \dots + 50) = 2 \cdot \frac{50 \cdot 51}{2} = 50 \cdot 51 = 2550
$$

Now, the total sum is:

$$
2500 - 2550 = -50
$$

---

### Step 3: General Formula for Alternating Series

For an alternating series of the form:

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

- If $ n $ is even, the sum is $ -\frac{n}{2} $
- If $ n $ is odd, the sum is $ \frac{n + 1}{2} $

In our case, $ n = 100 $, which is even, 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 nth odd number is 2n - 1. So if the last term is 99, then 2n - 1 = 99 => 2n = 100 => n = 50. So there are 50 terms here.

I remember that the sum of the first m odd numbers is m². Let me verify that. For example, sum of first 1 odd number: 1 = 1². First 2: 1 + 3 = 4 = 2². First 3: 1 + 3 + 5 = 9 = 3². Yep, that formula works. So the sum of the first 50 odd numbers 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).

The sum inside the parentheses is the sum of the first 50 even numbers. Because 2, 4, ..., 100. The nth even number is 2n. So the sum from n=1 to 50 of 2n.

Which is 2*(1 + 2 + 3 + ... + 50). The sum of the first m integers is m(m + 1)/2. So here, m = 50. Therefore, the sum is 2*(50*51)/2. The 2 and the denominator 2 cancel out, so it's 50*51 = 2550.

Therefore, the negative terms sum to -2550.

So the total sum is the sum of positive terms (2500) plus the sum of negative terms (-2550).

So 2500 - 2550 = -50.

Same result as before! So that confirms that the answer is indeed -50.

Alternatively, maybe I can use another formula. Let me recall that for an alternating series like this, there might be a general formula.

But since I already have two methods giving me -50, maybe I can be confident. However, let me try another approach just to be thorough.

Another way: Let me consider the series S = 1 - 2 + 3 - 4 + 5 - ... + 99 - 100.

If I write this as S = Σ_{k=1}^{100} (-1)^{k+1} * k.

Alternatively, I can think of this as S = Σ_{k=1}^{100} (-1)^{k+1} * k.

But maybe I can relate this to the formula for the sum of an alternating series. However, I don't remember a specific formula for this, but perhaps I can derive it.

Alternatively, consider that for an alternating series with terms alternating sign, if the number of terms is even, then we can pair them as (1 - 2) + (3 - 4) + ... + (99 - 100) as before, which gives -1 * (number of pairs). Since there are 100 terms, 50 pairs, so -50.

Alternatively, if I think of the general formula for the sum of such a series. Let me think.

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

If n is even, say n = 2m, then S = Σ_{k=1}^{2m} (-1)^{k+1} k.

Which can be written as Σ_{j=1}^{m} [ (2j - 1) - 2j ] = Σ_{j=1}^{m} (-1) = -m.

Since n = 2m, m = n/2. Therefore, S = -n/2.

In our case, n = 100, so S = -100/2 = -50.

Exactly! So that's another way to see it. Therefore, the answer is -50.

So all methods lead to the same conclusion.

But just to make sure, let me compute the sum step by step for a slightly larger number and see if the pattern holds.

For example, let's take n = 4: 1 - 2 + 3 - 4 = (1 - 2) + (3 - 4) = -1 + -1 = -2. According to the formula, -n/2 = -4/2 = -2. Correct.

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

So the formula seems to hold. Therefore, for n = 100, it's -100/2 = -50.

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

But just to make sure, let me try to compute the sum using another method. Let me consider the general formula for the sum of an alternating series.

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

This can be written as S = Σ_{k=1}^n (-1)^{k+1} k.

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

If we consider the sum S = Σ_{k=1}^n (-1)^{k+1} k.

Let me consider two cases: when n is even and when n is odd.

Case 1: n is even. Let n = 2m. Then S = Σ_{k=1}^{2m} (-1)^{k+1} k = Σ_{j=1}^m [ (2j - 1) - 2j ] = Σ_{j=1}^m (-1) = -m.

Which is -n/2.

Case 2: n is odd. Let n = 2m + 1. Then S = Σ_{k=1}^{2m+1} (-1)^{k+1} k = Σ_{j=1}^m [ (2j - 1) - 2j ] + (2m + 1) = Σ_{j=1}^m (-1) + (2m + 1) = -m + 2m + 1 = m + 1.

Which is (n + 1)/2.

So in our problem, n = 100, which is even, so S = -n/2 = -50.

Therefore, this formula confirms again that the answer is -50.

Alternatively, if I didn't remember the formula, I could derive it.

Let me try to derive it. Let me consider S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me write S as follows:

If n is even, then S = (1 - 2) + (3 - 4) + ... + (n-1 - n).

Each pair is -1, and there are n/2 pairs. Therefore, S = -n/2.

If n is odd, then S = (1 - 2) + (3 - 4) + ... + (n-2 - (n-1)) + n.

Each of the first (n-1)/2 pairs is -1, and then the last term is n. Therefore, S = - (n - 1)/2 + n = (-n + 1)/2 + n = (n + 1)/2.

So again, same result.

Therefore, since our problem has n = 100, which is even, the answer is -100/2 = -50.

Another way to think about it is using recursion or recurrence relations. Let me try that.

Let me denote S(n) as the sum up to n terms.

If n = 1, S(1) = 1.

If n = 2, S(2) = 1 - 2 = -1.

If n = 3, S(3) = 1 - 2 + 3 = 2.

If n = 4, S(4) = 1 - 2 + 3 - 4 = -2.

If n = 5, S(5) = 1 - 2 + 3 - 4 + 5 = 3.

So, it seems that for even n, S(n) = -n/2, and for odd n, S(n) = (n + 1)/2.

So for n = 100, which is even, S(100) = -100/2 = -50.

Therefore, this pattern also gives the same answer.

Alternatively, maybe I can use generating functions or other advanced methods, but that might be overkill.

Alternatively, think of the series as the difference between the sum of odd numbers and the sum of even numbers.

As I did earlier, the sum of odd numbers from 1 to 99 is 50² = 2500. The sum of even numbers from 2 to 100 is 2 + 4 + ... + 100. Which is 2*(1 + 2 + ... + 50) = 2*(50*51)/2 = 50*51 = 2550. Therefore, the total sum is 2500 - 2550 = -50.

Same result again.

So, all methods converge to -50.

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

But just to make sure, let me compute the sum manually for a small n and see.

Take n = 2: 1 - 2 = -1. According to formula, -2/2 = -1. Correct.

n = 4: 1 - 2 + 3 - 4 = -2. Formula: -4/2 = -2. Correct.

n = 6: 1 - 2 + 3 - 4 + 5 - 6 = -3. Formula: -6/2 = -3. Correct.

n = 100: -100/2 = -50.

Therefore, the answer is definitely -50.

Alternatively, if I think about the series as telescoping or something else? Not sure. But since all methods give the same answer, I think that's solid.

Another way: Let me consider the general term.

Let me write the series as S = Σ_{k=1}^{100} (-1)^{k+1} k.

Let me consider the sum S = Σ_{k=1}^{n} (-1)^{k+1} k.

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

If I consider the sum S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

This is an alternating series.

Let me consider the formula for the sum.

Let me recall that for an alternating series, the sum can be expressed as:

If n is even: S = -n/2

If n is odd: S = (n + 1)/2

Which is exactly what we derived earlier.

Therefore, since n = 100 is even, S = -100/2 = -50.

Therefore, the answer is -50.

Alternatively, if I didn't remember the formula, I could derive it using the method of differences.

Let me consider S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me consider multiplying S by -1:

- S = -1 + 2 - 3 + 4 - ... + (-1)^{n} n.

Now, if I add S and -S:

S + (-S) = 0 = [1 - 2 + 3 - 4 + ... + (-1)^{n+1} n] + [-1 + 2 - 3 + 4 - ... + (-1)^{n} n]

Let me compute term by term.

First term: 1 + (-1) = 0

Second term: -2 + 2 = 0

Third term: 3 + (-3) = 0

Fourth term: -4 + 4 = 0

...

Continuing this way, each pair cancels out.

But wait, how many terms are there? If n is even, then the last term in S is (-1)^{n+1} n = (-1)^{even + 1} n = (-1)^{odd} n = -n.

In -S, the last term is (-1)^{n} n = (-1)^{even} n = +n.

Therefore, adding S and -S:

All terms cancel except for the last term of S and the last term of -S? Wait, no. Wait, let me check again.

Wait, if n is even, then S has terms up to (-1)^{n+1} n = -n.

And -S has terms up to (-1)^{n} n = +n.

But when adding S and -S, each term cancels except for the last term? Wait, no. Let me think again.

Wait, if n is even, then the number of terms is even. Let me take n = 4 as an example.

S = 1 - 2 + 3 - 4

- S = -1 + 2 - 3 + 4

Adding them: (1 - 1) + (-2 + 2) + (3 - 3) + (-4 + 4) = 0 + 0 + 0 + 0 = 0.

So all terms cancel. Therefore, 0 = S + (-S) = 0. Which is trivial.

But how does that help me?

Alternatively, maybe I need to consider another approach.

Alternatively, let me consider S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me consider S = Σ_{k=1}^n (-1)^{k+1} k.

Let me consider the sum S + S.

Wait, maybe not. Alternatively, let me consider S = Σ_{k=1}^n (-1)^{k+1} k.

Let me write S as:

S = 1 - 2 + 3 - 4 + ... + (-1)^{n+1} n.

Let me consider S = Σ_{k=1}^n (-1)^{k+1} k.

Let me consider the difference between S and the sum of all numbers from 1 to n.

Wait, not sure.

Alternatively, think of S as the difference between the sum of odd terms and the sum of even terms.

Which is exactly what I did earlier.

Sum of odd terms: 1 + 3 + 5 + ... + 99 = 2500

Sum of even terms: 2 + 4 + 6 + ... + 100 = 2550

Therefore, S = 2500 - 2550 = -50.

Therefore, same result.

Alternatively, if I didn't remember the formula for the sum of odd or even numbers, I could compute them using arithmetic series formulas.

Sum of odd numbers: first term a1 = 1, last term an = 99, number of terms m = 50.

Sum = m*(a1 + an)/2 = 50*(1 + 99)/2 = 50*100/2 = 50*50 = 2500.

Sum of even numbers: first term a1 = 2, last term an = 100, number of terms m = 50.

Sum = m*(a1 + an)/2 = 50*(2 + 100)/2 = 50*102/2 = 50*51 = 2550.

Therefore, difference is 2500 - 2550 = -50.

Same result.

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

Therefore, I can be confident that the answer is \boxed{-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. This pattern continues up to the 100th term, which is negative.

---

### Step 1: Group the Terms in Pairs

We can group the terms in pairs to simplify the computation:

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

Each pair has the form $ (2k - 1) - 2k = -1 $, and since there are 100 terms, there are $ \frac{100}{2} = 50 $ such pairs.

So the total sum becomes:

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

---

### Step 2: Confirm Using Sum of Odd and Even Numbers

We can also compute the sum by separating the positive and negative terms:

- **Positive terms (odd numbers):** $ 1 + 3 + 5 + \dots + 99 $
- **Negative terms (even numbers):** $ -2 - 4 - 6 - \dots - 100 $

#### Sum of the first 50 odd numbers:

The sum of the first $ n $ odd numbers is $ n^2 $. Here, $ n = 50 $, so:

$$
1 + 3 + 5 + \dots + 99 = 50^2 = 2500
$$

#### Sum of the first 50 even numbers:

The sum of the first $ n $ even numbers is $ n(n + 1) $. Here, $ n = 50 $, so:

$$
2 + 4 + 6 + \dots + 100 = 2(1 + 2 + \dots + 50) = 2 \cdot \frac{50 \cdot 51}{2} = 50 \cdot 51 = 2550
$$

Now, the total sum is:

$$
2500 - 2550 = -50
$$

---

### Step 3: General Formula for Alternating Series

For an alternating series of the form:

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

- If $ n $ is even, the sum is $ -\frac{n}{2} $
- If $ n $ is odd, the sum is $ \frac{n + 1}{2} $

In our case, $ n = 100 $, which is even, so:

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

---

### Final Answer

$$
\boxed{-50}
$$

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