answer：Okay, so I have this sequence problem here. The sequence is defined by the recurrence relation ((n - 1)a_n = (n + 1)a_{n-1}), and they give me that (a_1 = 1). I need to find (a_{100}). Hmm, let me try to figure this out step by step.First, let me write down the recurrence relation again to make sure I have it correctly:[(n - 1)a_n = (n + 1)a_{n-1}]So, for each term (a_n), it's related to the previous term (a_{n-1}) by this equation. Maybe I can solve for (a_n) in terms of (a_{n-1}). Let me do that.Divide both sides by ((n - 1)):[a_n = frac{n + 1}{n - 1} a_{n-1}]Okay, so each term is (frac{n + 1}{n - 1}) times the previous term. That seems like a ratio that I can perhaps telescope when I write out the terms.Let me try writing out the first few terms to see if I can spot a pattern.Given (a_1 = 1), let's compute (a_2):For (n = 2):[a_2 = frac{2 + 1}{2 - 1} a_{1} = frac{3}{1} times 1 = 3]So, (a_2 = 3).Now, (a_3):For (n = 3):[a_3 = frac{3 + 1}{3 - 1} a_{2} = frac{4}{2} times 3 = 2 times 3 = 6]So, (a_3 = 6).Next, (a_4):For (n = 4):[a_4 = frac{4 + 1}{4 - 1} a_{3} = frac{5}{3} times 6 = frac{5 times 6}{3} = 5 times 2 = 10]So, (a_4 = 10).Wait a second, these numbers look familiar. 1, 3, 6, 10... These are triangular numbers! The nth triangular number is given by (frac{n(n + 1)}{2}). Let me check if this holds.For (n = 1): (frac{1 times 2}{2} = 1), which matches (a_1).For (n = 2): (frac{2 times 3}{2} = 3), which matches (a_2).For (n = 3): (frac{3 times 4}{2} = 6), which matches (a_3).For (n = 4): (frac{4 times 5}{2} = 10), which matches (a_4).So, it seems like (a_n = frac{n(n + 1)}{2}). But wait, let me verify this with the recurrence relation to make sure.Assume that (a_{n-1} = frac{(n - 1)n}{2}). Then, according to the recurrence:[a_n = frac{n + 1}{n - 1} a_{n-1} = frac{n + 1}{n - 1} times frac{(n - 1)n}{2} = frac{(n + 1)n}{2}]Yes, that simplifies to (frac{n(n + 1)}{2}), which is exactly the formula for the nth triangular number. So, it seems my initial guess is correct.Therefore, the general term is (a_n = frac{n(n + 1)}{2}). Therefore, (a_{100}) would be:[a_{100} = frac{100 times 101}{2} = frac{10100}{2} = 5050]Wait, let me double-check my calculations to be sure.Compute (100 times 101):100 multiplied by 100 is 10,000, and 100 multiplied by 1 is 100, so 10,000 + 100 = 10,100.Divide that by 2: 10,100 / 2 = 5,050.Yes, that seems right.But just to be thorough, let me try computing (a_5) using both the recurrence and the formula to see if they match.Using the recurrence:(a_4 = 10), so for (n = 5):[a_5 = frac{5 + 1}{5 - 1} a_4 = frac{6}{4} times 10 = frac{3}{2} times 10 = 15]Using the formula:(a_5 = frac{5 times 6}{2} = frac{30}{2} = 15)They match. So, my formula is correct.Therefore, I can confidently say that (a_{100} = 5050).**Final Answer**The value of (a_{100}) is boxed{5050}.

answer：Okay, so I have this problem where a person needs to exercise for at least 60 minutes in a day. There are five types of exercises they can choose from, each with different durations. The exercises are labeled A to E, and each has a specific time slot and duration. The goal is to figure out how many different ways they can combine these exercises to meet or exceed the 60-minute requirement.First, let me list out the details of each exercise to make sure I have them straight:- **Exercise A**: 7:00 - 8:00, 30 minutes- **Exercise B**: 8:00 - 9:00, 20 minutes- **Exercise C**: 9:00 - 10:00, 40 minutes- **Exercise D**: 10:00 - 11:00, 30 minutes- **Exercise E**: 11:00 - 12:00, 30 minutesLooking at the time slots, each exercise is in a different hour block, and none of them overlap. That means the person can potentially do multiple exercises in a day without worrying about time conflicts. However, each exercise has a fixed duration, so the total time will depend on how many exercises they choose and which ones they pick.The person needs at least 60 minutes of exercise. So, the total duration of the chosen exercises should be ≥60 minutes. Since each exercise is non-overlapping, the order in which they are done doesn't affect the total duration, only the combination of exercises.Now, let's think about how to approach this. It seems like a combinatorial problem where we need to count the number of subsets of the exercises whose total duration is at least 60 minutes. Each exercise can either be chosen or not, so we're dealing with combinations here.Let me list the durations again for clarity:- A: 30- B: 20- C: 40- D: 30- E: 30So, the durations are: 30, 20, 40, 30, 30.We need to find all possible subsets of these durations where the sum is ≥60.To solve this, I can consider all possible combinations of the exercises and count those that meet the duration requirement. Since there are 5 exercises, the total number of possible subsets is 2^5 = 32. However, not all of these subsets will meet the 60-minute requirement. So, I need to find how many of these 32 subsets have a total duration of 60 or more minutes.Alternatively, instead of listing all 32 subsets, which might be time-consuming, I can approach this systematically by considering the number of exercises chosen and calculating the possible durations for each case.Let's break it down by the number of exercises:1. **Choosing 1 exercise**: The maximum duration here is 40 minutes (Exercise C). Since 40 < 60, none of the single-exercise subsets will meet the requirement.2. **Choosing 2 exercises**: Let's find all possible pairs and their durations. - A + B: 30 + 20 = 50 - A + C: 30 + 40 = 70 - A + D: 30 + 30 = 60 - A + E: 30 + 30 = 60 - B + C: 20 + 40 = 60 - B + D: 20 + 30 = 50 - B + E: 20 + 30 = 50 - C + D: 40 + 30 = 70 - C + E: 40 + 30 = 70 - D + E: 30 + 30 = 60 Now, let's count how many of these pairs meet or exceed 60 minutes: - A + C: 70 ✔️ - A + D: 60 ✔️ - A + E: 60 ✔️ - B + C: 60 ✔️ - C + D: 70 ✔️ - C + E: 70 ✔️ - D + E: 60 ✔️ So, that's 7 pairs.3. **Choosing 3 exercises**: Let's consider all triplets and their durations. - A + B + C: 30 + 20 + 40 = 90 - A + B + D: 30 + 20 + 30 = 80 - A + B + E: 30 + 20 + 30 = 80 - A + C + D: 30 + 40 + 30 = 100 - A + C + E: 30 + 40 + 30 = 100 - A + D + E: 30 + 30 + 30 = 90 - B + C + D: 20 + 40 + 30 = 90 - B + C + E: 20 + 40 + 30 = 90 - B + D + E: 20 + 30 + 30 = 80 - C + D + E: 40 + 30 + 30 = 100 All of these triplets have durations ≥60 minutes. There are 10 triplets in total.4. **Choosing 4 exercises**: Let's look at all quadruplets. - A + B + C + D: 30 + 20 + 40 + 30 = 120 - A + B + C + E: 30 + 20 + 40 + 30 = 120 - A + B + D + E: 30 + 20 + 30 + 30 = 110 - A + C + D + E: 30 + 40 + 30 + 30 = 130 - B + C + D + E: 20 + 40 + 30 + 30 = 120 All of these quadruplets have durations ≥60 minutes. There are 5 quadruplets.5. **Choosing all 5 exercises**: The total duration is 30 + 20 + 40 + 30 + 30 = 150 minutes, which is way above 60. So, this is one valid combination.Now, let's sum up all the valid combinations:- 2 exercises: 7- 3 exercises: 10- 4 exercises: 5- 5 exercises: 1Adding these together: 7 + 10 + 5 + 1 = 23.Wait, hold on. That gives me 23, but let me double-check my counts because sometimes it's easy to miscount.Starting with 2 exercises:- A + C: 70 ✔️- A + D: 60 ✔️- A + E: 60 ✔️- B + C: 60 ✔️- C + D: 70 ✔️- C + E: 70 ✔️- D + E: 60 ✔️Yes, that's 7.For 3 exercises, all 10 combinations are valid because the minimum duration would be the smallest three durations: 20 + 30 + 30 = 80, which is above 60. So, 10 is correct.For 4 exercises, all 5 combinations are valid because the minimum duration would be 20 + 30 + 30 + 30 = 110, which is way above 60. So, 5 is correct.And the 5-exercise combination is obviously valid.So, 7 + 10 + 5 + 1 = 23.But wait, hold on a second. The problem says "at least 60 minutes." So, does that include exactly 60 or more? Yes, it does. So, our count includes all subsets where the total duration is 60 or more.But let me think again. Is 23 the correct total?Wait, another approach is to calculate the total number of subsets and subtract those that don't meet the requirement.Total subsets: 32.Subsets that don't meet the requirement: subsets with total duration <60.So, if I can calculate the number of subsets with total duration <60, then subtract that from 32 to get the answer.Let me try this approach to verify.First, subsets with 0 exercises: 1 subset (doing nothing). But since the person needs to exercise, I think we can exclude this. Wait, the problem says "at least 60 minutes," so doing nothing is not an option. So, the total subsets we're considering are from 1 to 5 exercises.But in the problem statement, it's not specified whether the person must do at least one exercise or can choose to do none. Hmm, the problem says "needs to exercise for a total of at least 60 minutes," so I think they have to do some exercises, so the empty set is excluded.So, total subsets: 32 -1 =31.Wait, no. Wait, the empty set is just one subset, but the person needs to exercise, so we're only considering subsets with at least one exercise. So, total subsets: 31.But wait, actually, no. The total number of subsets is 32, including the empty set. Since the person must exercise, we have 31 subsets to consider.But actually, in the problem statement, it's not clear whether the person must exercise or can choose not to. Hmm. The problem says "needs to exercise for a total of at least 60 minutes on a certain day." So, it implies that they must exercise, so the empty set is not considered. So, total subsets: 31.But in my initial approach, I considered all subsets, including the empty set, but since the person must exercise, we should only consider subsets with at least one exercise.But in my first calculation, I considered subsets of size 1 to 5, which is 31 subsets. So, that's correct.So, if I can calculate the number of subsets with total duration <60, then subtract that from 31 to get the number of subsets with total duration ≥60.Let me try that.First, subsets with 1 exercise: 5 subsets.Each single exercise has duration: 30, 20, 40, 30, 30.So, the durations are: 20, 30, 30, 30, 40.So, the single-exercise subsets with duration <60: all of them, since the maximum is 40 <60. So, 5 subsets.Subsets with 2 exercises: as before, 10 subsets.But how many of these have total duration <60?Looking back at the pairs:- A + B: 50 <60 ✔️- A + C: 70 ≥60 ❌- A + D: 60 ≥60 ❌- A + E: 60 ≥60 ❌- B + C: 60 ≥60 ❌- B + D: 50 <60 ✔️- B + E: 50 <60 ✔️- C + D: 70 ≥60 ❌- C + E: 70 ≥60 ❌- D + E: 60 ≥60 ❌So, the pairs with total duration <60 are:- A + B: 50- B + D: 50- B + E: 50That's 3 subsets.So, 3 subsets with 2 exercises have total duration <60.Subsets with 3 exercises: 10 subsets.What's the minimum duration for 3 exercises? The smallest three durations: 20 + 30 + 30 = 80, which is above 60. So, all 3-exercise subsets have duration ≥60. Therefore, no subsets with 3 exercises have duration <60.Similarly, subsets with 4 or 5 exercises will have even larger durations, so they all have duration ≥60.Therefore, the total number of subsets with duration <60 is:- 1-exercise: 5- 2-exercise: 3- 3-exercise: 0- 4-exercise: 0- 5-exercise: 0Total: 5 + 3 = 8 subsets.Therefore, the number of subsets with duration ≥60 is total subsets (31) minus subsets with duration <60 (8): 31 - 8 = 23.So, that confirms my initial count of 23.Wait, but hold on. Earlier, when I broke it down by the number of exercises, I had:- 2 exercises: 7- 3 exercises: 10- 4 exercises: 5- 5 exercises: 1Total: 23.Yes, that's consistent.So, both methods give me 23.But let me just make sure I didn't miss any subsets.Another way to think about it is to list all possible subsets and count those with total duration ≥60. But since there are 31 subsets, it's a bit tedious, but let me try to list them systematically.List of exercises: A(30), B(20), C(40), D(30), E(30).Let me list all subsets:1. A: 302. B: 203. C: 404. D: 305. E: 306. A+B:507. A+C:708. A+D:609. A+E:6010. B+C:6011. B+D:5012. B+E:5013. C+D:7014. C+E:7015. D+E:6016. A+B+C:9017. A+B+D:8018. A+B+E:8019. A+C+D:10020. A+C+E:10021. A+D+E:9022. B+C+D:9023. B+C+E:9024. B+D+E:8025. C+D+E:10026. A+B+C+D:12027. A+B+C+E:12028. A+B+D+E:11029. A+C+D+E:13030. B+C+D+E:12031. A+B+C+D+E:150Now, let's go through each subset and mark whether it's ≥60:1. A:30 - No2. B:20 - No3. C:40 - No4. D:30 - No5. E:30 - No6. A+B:50 - No7. A+C:70 - Yes8. A+D:60 - Yes9. A+E:60 - Yes10. B+C:60 - Yes11. B+D:50 - No12. B+E:50 - No13. C+D:70 - Yes14. C+E:70 - Yes15. D+E:60 - Yes16. A+B+C:90 - Yes17. A+B+D:80 - Yes18. A+B+E:80 - Yes19. A+C+D:100 - Yes20. A+C+E:100 - Yes21. A+D+E:90 - Yes22. B+C+D:90 - Yes23. B+C+E:90 - Yes24. B+D+E:80 - Yes25. C+D+E:100 - Yes26. A+B+C+D:120 - Yes27. A+B+C+E:120 - Yes28. A+B+D+E:110 - Yes29. A+C+D+E:130 - Yes30. B+C+D+E:120 - Yes31. A+B+C+D+E:150 - YesNow, let's count the "Yes" responses:From 7 to 31, let's count:7. Yes8. Yes9. Yes10. Yes13. Yes14. Yes15. Yes16. Yes17. Yes18. Yes19. Yes20. Yes21. Yes22. Yes23. Yes24. Yes25. Yes26. Yes27. Yes28. Yes29. Yes30. Yes31. YesThat's 23 subsets.So, this confirms it. There are 23 subsets where the total duration is at least 60 minutes.Therefore, the number of ways to combine the exercises is 23.**Final Answer**The number of ways to combine the exercises is boxed{23}.

answer：Okay, so I have this problem about a geometric sequence. It says that the sequence {a_n} is an increasing geometric sequence. Hmm, okay, so that means each term is bigger than the one before it. So, the common ratio q must be greater than 1, right? Because if q is less than 1, the sequence would be decreasing, and if q is 1, it's constant. So, q > 1.They give me two pieces of information: a₁ + a₄ = 9, and a₂a₃ = 8. I need to find the common ratio q.Let me recall what a geometric sequence is. Each term is the previous term multiplied by q. So, a₁ is the first term, a₂ = a₁*q, a₃ = a₂*q = a₁*q², a₄ = a₃*q = a₁*q³, and so on. So, in general, a_n = a₁*q^{n-1}.So, let's write down the given information in terms of a₁ and q.First, a₁ + a₄ = 9. Since a₄ is a₁*q³, that equation becomes:a₁ + a₁*q³ = 9.I can factor out a₁:a₁*(1 + q³) = 9. Let me write that as equation (1):(1) a₁*(1 + q³) = 9.Next, the product of a₂ and a₃ is 8. Let's write that out.a₂ = a₁*q, and a₃ = a₁*q². So, their product is:a₂*a₃ = (a₁*q)*(a₁*q²) = a₁²*q³ = 8.So, equation (2) is:(2) a₁²*q³ = 8.Now, I have two equations:1) a₁*(1 + q³) = 92) a₁²*q³ = 8I need to solve for q. So, maybe I can express a₁ from equation (1) and substitute into equation (2). Let's try that.From equation (1):a₁ = 9 / (1 + q³)Now, plug this into equation (2):(9 / (1 + q³))² * q³ = 8Let me write that out:(81) / (1 + q³)² * q³ = 8So, 81*q³ / (1 + q³)² = 8Hmm, that looks a bit complicated, but maybe I can simplify it.Let me denote t = q³. Since q > 1, t will be greater than 1 as well.So, substituting t into the equation:81*t / (1 + t)² = 8So, now we have:81t / (1 + t)² = 8Let me write that as:81t = 8*(1 + t)²Expanding the right-hand side:8*(1 + 2t + t²) = 8 + 16t + 8t²So, the equation becomes:81t = 8 + 16t + 8t²Let me bring all terms to one side:8t² + 16t + 8 - 81t = 0Simplify the terms:8t² - 65t + 8 = 0So, now I have a quadratic equation in terms of t:8t² - 65t + 8 = 0I can try to solve this quadratic for t. Let's use the quadratic formula.t = [65 ± sqrt(65² - 4*8*8)] / (2*8)Calculating discriminant D:D = 65² - 4*8*8 = 4225 - 256 = 3969sqrt(3969) is... let me see, 63² is 3969, because 60²=3600, 63²=3969. So, sqrt(D)=63.Therefore,t = [65 ± 63] / 16So, two solutions:t₁ = (65 + 63)/16 = 128/16 = 8t₂ = (65 - 63)/16 = 2/16 = 1/8So, t can be 8 or 1/8.But remember that t = q³, and since the sequence is increasing, q > 1, so t = q³ must be greater than 1. Therefore, t = 8 is acceptable, and t = 1/8 is not, because that would make q < 1, which would make the sequence decreasing, which contradicts the given information.Therefore, t = 8, so q³ = 8.Therefore, q = cube root of 8 = 2.So, the common ratio is 2.Let me just verify this to make sure I didn't make any mistakes.If q = 2, then let's find a₁.From equation (1):a₁*(1 + q³) = 9q³ = 8, so:a₁*(1 + 8) = 9 => a₁*9 = 9 => a₁ = 1.So, a₁ = 1, q = 2.Let's check the second condition: a₂*a₃ = 8.a₂ = a₁*q = 1*2 = 2a₃ = a₁*q² = 1*4 = 4So, a₂*a₃ = 2*4 = 8. That's correct.And the first condition: a₁ + a₄ = 1 + a₁*q³ = 1 + 8 = 9. Correct.So, everything checks out. Therefore, the common ratio q is 2.**Final Answer**The common ratio ( q ) is boxed{2}.

answer：Okay, let me try to figure out this problem step by step. It's a bit complex, but I'll take it slowly.First, let's summarize the given information:- Location A is 60 km west of location B. So, the distance between A and B is 60 km.- Individuals A and B start from A, while individuals C and D start from B at the same time.- A, B, and D are traveling east, while C is traveling west.- The speeds of A, B, C, and D form an arithmetic sequence, with A being the fastest.- After n hours, B and C meet.- Then, n hours later, A catches up to D at location C.- We need to find the distance between locations B and C.Alright, let's break this down.First, let's denote the speeds of A, B, C, D. Since their speeds form an arithmetic sequence and A is the fastest, let's denote their speeds as follows:Let the common difference be d. So, if we let the speed of A be a, then the speeds of B, C, D will be a - d, a - 2d, a - 3d respectively. Wait, but since it's an arithmetic sequence, and A is the highest, the next ones should be decreasing by d each time.But hold on, we have four individuals: A, B, C, D. So, the speeds are in an arithmetic sequence with A being the highest. So, the order is A, B, C, D with speeds decreasing by d each time. So, speed of A is a, speed of B is a - d, speed of C is a - 2d, and speed of D is a - 3d.But wait, let me make sure. The problem says "the speeds of A, B, C, and D form an arithmetic sequence, with A having the highest speed." So, the order is A, B, C, D, each subsequent speed is lower by a common difference d. So, yes, speed of A = a, speed of B = a - d, speed of C = a - 2d, speed of D = a - 3d.Alright, so we have:- Speed of A: a- Speed of B: a - d- Speed of C: a - 2d- Speed of D: a - 3dAll right, now let's think about their movements.Individuals A and B start from A, which is 60 km west of B. So, A is at location A, and B is also at location A? Wait, no. Wait, A is a location, and individuals A and B are departing from location A. So, both A and B start from A, which is 60 km west of B. So, location B is 60 km east of A.Similarly, individuals C and D start from location B.So, to clarify:- Location A is 60 km west of location B.- A and B start from A, moving east.- C and D start from B, with C moving west and D moving east.So, A and B are moving east from A, which is 60 km west of B. C is moving west from B, and D is moving east from B.Got it.Now, after n hours, B and C meet. Then, n hours later, A catches up to D at location C.Wait, so first, B and C meet after n hours. Then, n hours after that meeting, A catches up to D at location C.Wait, but location C is where they meet? Or is location C a different place? Wait, the problem says "A catches up to D at location C." So, location C is where A catches up to D, which is n hours after B and C met.Wait, but initially, C started from B, moving west. So, location C is somewhere west of B, right? Because C is moving west from B.But when A catches up to D at location C, that must mean that location C is a point where both A and D are present after some time.Wait, this is getting a bit confusing. Let me try to visualize it.Let me draw a number line with location A at 0 and location B at 60 km.So, A is at 0, B is at 60.Individuals:- A: starts at 0, moving east (towards positive direction) at speed a.- B: starts at 0, moving east at speed a - d.- C: starts at 60, moving west (towards negative direction) at speed a - 2d.- D: starts at 60, moving east at speed a - 3d.So, after n hours, B and C meet. Let's find where they meet.In n hours, B has moved a distance of (a - d) * n from A (which is at 0). So, B's position is (a - d) * n.Similarly, C has moved west from B (which is at 60) towards A. So, C's position is 60 - (a - 2d) * n.Since they meet, their positions are equal:(a - d) * n = 60 - (a - 2d) * nLet me write that equation:(a - d) * n = 60 - (a - 2d) * nLet me solve for n.Bring all terms to one side:(a - d) * n + (a - 2d) * n - 60 = 0Factor n:[(a - d) + (a - 2d)] * n - 60 = 0Simplify inside the brackets:(a - d + a - 2d) = 2a - 3dSo:(2a - 3d) * n - 60 = 0Thus:(2a - 3d) * n = 60So,n = 60 / (2a - 3d)Alright, that's the first equation.Now, after this meeting, n hours later, A catches up to D at location C.Wait, so the total time from the start is n + n = 2n hours when A catches up to D.But wait, the problem says: "After n hours, B and C meet, and n hours later, A catches up to D at location C."So, the first event is at n hours: B and C meet.Then, from that point, another n hours pass, so total time is 2n hours when A catches up to D at location C.So, at time 2n, A and D are at location C.So, let's find where A and D are at time 2n.First, A's position at time 2n:A is moving east from A (0) at speed a, so position is a * 2n.D's position at time 2n:D is moving east from B (60) at speed a - 3d, so position is 60 + (a - 3d) * 2n.Since they meet at location C, their positions are equal:a * 2n = 60 + (a - 3d) * 2nLet me write that equation:2a * n = 60 + 2n(a - 3d)Simplify the right side:60 + 2a * n - 6d * nSo, equation becomes:2a * n = 60 + 2a * n - 6d * nSubtract 2a * n from both sides:0 = 60 - 6d * nSo,6d * n = 60Thus,d * n = 10So,d = 10 / nAlright, so we have d in terms of n.Earlier, we had n = 60 / (2a - 3d)So, let's substitute d = 10 / n into this equation.n = 60 / (2a - 3*(10 / n))Simplify denominator:2a - 30 / nSo,n = 60 / (2a - 30 / n)Multiply numerator and denominator by n to eliminate the fraction:n = (60n) / (2a * n - 30)So,n * (2a * n - 30) = 60nDivide both sides by n (assuming n ≠ 0, which it isn't since they meet after some time):2a * n - 30 = 60Thus,2a * n = 90So,a * n = 45So, a = 45 / nAlright, so now we have:a = 45 / nandd = 10 / nSo, let's note these down.Now, let's recall the speeds:- Speed of A: a = 45 / n- Speed of B: a - d = (45 / n) - (10 / n) = 35 / n- Speed of C: a - 2d = (45 / n) - 20 / n = 25 / n- Speed of D: a - 3d = (45 / n) - 30 / n = 15 / nSo, the speeds are:A: 45/n, B: 35/n, C:25/n, D:15/n.So, the speeds are in arithmetic sequence with common difference d = 10/n.Alright, that seems consistent.Now, let's think about the positions.We know that at time n, B and C meet.So, position of B at time n is (35/n) * n = 35 km.Similarly, position of C at time n is 60 - (25/n) * n = 60 - 25 = 35 km.So, they meet at 35 km from A, which is 60 - 35 = 25 km west of B.Wait, so location where B and C meet is 35 km east of A, which is 25 km west of B.So, that's the meeting point.Then, n hours later, so at time 2n, A catches up to D at location C.Wait, but location C is where they meet, which is 35 km from A.Wait, no. Wait, the problem says "A catches up to D at location C." So, location C is where A and D meet, which is at time 2n.But earlier, B and C met at location 35 km east of A, which is 25 km west of B.So, is location C the same as the meeting point of B and C? Or is it a different location?Wait, the problem says "n hours later, A catches up to D at location C." So, location C is a specific place, which is where A and D meet. So, perhaps location C is a different point.Wait, but in the problem statement, location C is mentioned as a starting point for individual C. Wait, no, individuals C and D start from location B.Wait, maybe location C is a point somewhere on the route, not necessarily the starting point.Wait, the problem says "A catches up to D at location C." So, location C is a point where A and D meet. So, perhaps it's a different location from where B and C met.Wait, but in the problem statement, it's just mentioned as location C, which might be a specific point.Wait, perhaps I need to clarify.Wait, let me re-read the problem:"Individuals A and B depart from location A while individuals C and D depart from location B at the same time. A, B, and D all travel east, while C travels west. It is known that the speeds of A, B, C, and D form an arithmetic sequence, with A having the highest speed. After n hours, B and C meet, and n hours later, A catches up to D at location C. What is the distance between locations B and C in kilometers?"So, location C is where A catches up to D, which is n hours after B and C met.So, location C is a point on the route where A and D meet, which is n hours after the meeting of B and C.So, location C is a different point from where B and C met.So, let's figure out where location C is.At time 2n, A has traveled a * 2n = (45/n) * 2n = 90 km.Similarly, D has traveled from B (60 km east of A) eastward at speed 15/n for 2n hours, so D's position is 60 + (15/n)*2n = 60 + 30 = 90 km.So, both A and D are at 90 km from A, which is 90 - 60 = 30 km east of B.Wait, so location C is 90 km from A, which is 30 km east of B.Therefore, the distance between B and C is 30 km.Wait, so the answer is 30 km, which is option B.But let me verify this step by step to make sure.So, let's recap:- Speeds: A=45/n, B=35/n, C=25/n, D=15/n.- After n hours, B has moved 35/n * n = 35 km from A.- C has moved 25/n * n = 25 km west from B, so position is 60 -25=35 km from A.- So, they meet at 35 km from A, which is 25 km west of B.- Then, n hours later (total time 2n), A has moved 45/n * 2n=90 km.- D has moved from B (60 km) eastward at 15/n * 2n=30 km, so position is 60+30=90 km.- So, A and D meet at 90 km from A, which is 30 km east of B.- Therefore, location C is 30 km east of B.- So, the distance between B and C is 30 km.Therefore, the answer is 30 km, which is option B.Wait, but let me check if I interpreted location C correctly.In the problem statement, it says "A catches up to D at location C." So, location C is where A and D meet, which is 90 km from A, 30 km east of B.Therefore, the distance between B and C is 30 km.Yes, that seems correct.So, the answer is 30 km, which is option B.**Final Answer**boxed{B}

answer：Okay, so I have this problem where I need to find the maximum value of the expression (sin(alpha - beta) + 2 sin(alpha + beta)), where both (alpha) and (beta) are in the interval ([0, frac{pi}{4}]). Hmm, let me think about how to approach this.First, I remember that when dealing with trigonometric expressions, sometimes it's helpful to use trigonometric identities to simplify them. Maybe I can rewrite the expression in terms of sine and cosine of (alpha) and (beta) separately. Let me recall the sine addition and subtraction formulas.The sine of a difference is (sin(alpha - beta) = sinalpha cosbeta - cosalpha sinbeta), and the sine of a sum is (sin(alpha + beta) = sinalpha cosbeta + cosalpha sinbeta). So, substituting these into the original expression, I get:[sin(alpha - beta) + 2 sin(alpha + beta) = [sinalpha cosbeta - cosalpha sinbeta] + 2[sinalpha cosbeta + cosalpha sinbeta]]Let me expand this:[= sinalpha cosbeta - cosalpha sinbeta + 2sinalpha cosbeta + 2cosalpha sinbeta]Now, combine like terms:- The terms with (sinalpha cosbeta): (1 + 2 = 3), so (3sinalpha cosbeta)- The terms with (cosalpha sinbeta): (-1 + 2 = 1), so (cosalpha sinbeta)So, the expression simplifies to:[3sinalpha cosbeta + cosalpha sinbeta]Hmm, that seems a bit simpler. Maybe I can factor this expression further or express it in terms of a single trigonometric function. Let me see.Alternatively, perhaps I can consider treating this as a function of two variables, (f(alpha, beta) = 3sinalpha cosbeta + cosalpha sinbeta), and find its maximum over the domain (alpha, beta in [0, frac{pi}{4}]).To find the maximum, I can use calculus. Specifically, I can find the critical points by taking partial derivatives with respect to (alpha) and (beta), set them equal to zero, and solve for (alpha) and (beta). Then, I can check the boundary points as well since the maximum could occur there.Let me compute the partial derivatives.First, the partial derivative with respect to (alpha):[frac{partial f}{partial alpha} = 3cosalpha cosbeta - sinalpha sinbeta]And the partial derivative with respect to (beta):[frac{partial f}{partial beta} = -3sinalpha sinbeta + cosalpha cosbeta]To find critical points, set both partial derivatives equal to zero:1. (3cosalpha cosbeta - sinalpha sinbeta = 0)2. (-3sinalpha sinbeta + cosalpha cosbeta = 0)Hmm, so we have a system of two equations:Equation (1): (3cosalpha cosbeta = sinalpha sinbeta)Equation (2): (-3sinalpha sinbeta = -cosalpha cosbeta), which simplifies to (3sinalpha sinbeta = cosalpha cosbeta)Wait, that's interesting. Equation (2) is actually the same as Equation (1). Let me check:From Equation (1): (3cosalpha cosbeta = sinalpha sinbeta)From Equation (2): (3sinalpha sinbeta = cosalpha cosbeta)So, if I substitute Equation (1) into Equation (2):From Equation (1): (sinalpha sinbeta = 3cosalpha cosbeta)Substitute into Equation (2):(3(3cosalpha cosbeta) = cosalpha cosbeta)Simplify:(9cosalpha cosbeta = cosalpha cosbeta)Subtract (cosalpha cosbeta) from both sides:(8cosalpha cosbeta = 0)So, either (cosalpha = 0) or (cosbeta = 0). But since (alpha, beta in [0, frac{pi}{4}]), (cosalpha) and (cosbeta) are both positive and not zero. Therefore, there are no critical points inside the domain where both partial derivatives are zero.That means the maximum must occur on the boundary of the domain. So, I need to check the boundaries where either (alpha = 0), (alpha = frac{pi}{4}), (beta = 0), or (beta = frac{pi}{4}).Let me consider each case.**Case 1: (alpha = 0)**Then, the expression becomes:(3sin(0)cosbeta + cos(0)sinbeta = 0 + 1 cdot sinbeta = sinbeta)Since (beta in [0, frac{pi}{4}]), the maximum of (sinbeta) is at (beta = frac{pi}{4}), which is (sinfrac{pi}{4} = frac{sqrt{2}}{2} approx 0.707)**Case 2: (alpha = frac{pi}{4})**The expression becomes:(3sinleft(frac{pi}{4}right)cosbeta + cosleft(frac{pi}{4}right)sinbeta)We know that (sinleft(frac{pi}{4}right) = cosleft(frac{pi}{4}right) = frac{sqrt{2}}{2}), so:(3 cdot frac{sqrt{2}}{2} cosbeta + frac{sqrt{2}}{2} sinbeta = frac{3sqrt{2}}{2} cosbeta + frac{sqrt{2}}{2} sinbeta)Factor out (frac{sqrt{2}}{2}):(frac{sqrt{2}}{2} (3cosbeta + sinbeta))Now, to find the maximum of this expression over (beta in [0, frac{pi}{4}]). Let's denote (g(beta) = 3cosbeta + sinbeta). We can find its maximum by taking its derivative.(g'(beta) = -3sinbeta + cosbeta)Set derivative equal to zero:(-3sinbeta + cosbeta = 0)(cosbeta = 3sinbeta)Divide both sides by (cosbeta) (assuming (cosbeta neq 0)):(1 = 3tanbeta)(tanbeta = frac{1}{3})So, (beta = arctanleft(frac{1}{3}right)). Let me compute (arctanleft(frac{1}{3}right)). Since (frac{1}{3}) is approximately 0.333, and (arctan(0.333)) is approximately 0.32175 radians, which is about 18.43 degrees. Since (frac{pi}{4}) is approximately 0.7854 radians (45 degrees), this critical point is within our interval [0, (frac{pi}{4})].So, we need to evaluate (g(beta)) at (beta = arctanleft(frac{1}{3}right)), as well as at the endpoints (beta = 0) and (beta = frac{pi}{4}).First, compute (gleft(arctanleft(frac{1}{3}right)right)):Let me denote (beta = arctanleft(frac{1}{3}right)). Then, (sinbeta = frac{1}{sqrt{10}}) and (cosbeta = frac{3}{sqrt{10}}). Let me verify:If (tanbeta = frac{1}{3}), then the opposite side is 1, adjacent side is 3, hypotenuse is (sqrt{1^2 + 3^2} = sqrt{10}). So yes, (sinbeta = frac{1}{sqrt{10}}), (cosbeta = frac{3}{sqrt{10}}).So,(g(beta) = 3 cdot frac{3}{sqrt{10}} + frac{1}{sqrt{10}} = frac{9}{sqrt{10}} + frac{1}{sqrt{10}} = frac{10}{sqrt{10}} = sqrt{10})So, (g(beta)) at the critical point is (sqrt{10}), which is approximately 3.1623.Now, evaluate (g(beta)) at (beta = 0):(g(0) = 3cos(0) + sin(0) = 3 cdot 1 + 0 = 3)At (beta = frac{pi}{4}):(gleft(frac{pi}{4}right) = 3cosleft(frac{pi}{4}right) + sinleft(frac{pi}{4}right) = 3 cdot frac{sqrt{2}}{2} + frac{sqrt{2}}{2} = frac{3sqrt{2}}{2} + frac{sqrt{2}}{2} = 2sqrt{2} approx 2.8284)So, the maximum of (g(beta)) is (sqrt{10}) at (beta = arctanleft(frac{1}{3}right)). Therefore, the expression when (alpha = frac{pi}{4}) is:(frac{sqrt{2}}{2} cdot sqrt{10} = frac{sqrt{20}}{2} = frac{2sqrt{5}}{2} = sqrt{5} approx 2.2361)Wait, hold on, that doesn't seem right. Wait, no, actually, let me compute that again.Wait, (g(beta)) is multiplied by (frac{sqrt{2}}{2}). So, (g(beta) = sqrt{10}), so the expression is (frac{sqrt{2}}{2} times sqrt{10} = frac{sqrt{20}}{2} = frac{2sqrt{5}}{2} = sqrt{5}). So, yes, that's correct.But wait, (sqrt{5}) is approximately 2.236, which is less than (sqrt{10}), but we already considered that in the expression.Wait, but actually, the expression when (alpha = frac{pi}{4}) is (frac{sqrt{2}}{2} times g(beta)), so the maximum is (sqrt{5}), which is approximately 2.236.But earlier, in Case 1, when (alpha = 0), the maximum was about 0.707, which is much less. So, so far, the maximum is (sqrt{5}).Wait, but hold on, let me check the other cases.**Case 3: (beta = 0)**The expression becomes:(3sinalpha cos(0) + cosalpha sin(0) = 3sinalpha cdot 1 + cosalpha cdot 0 = 3sinalpha)Since (alpha in [0, frac{pi}{4}]), the maximum of (3sinalpha) occurs at (alpha = frac{pi}{4}), which is (3 cdot frac{sqrt{2}}{2} = frac{3sqrt{2}}{2} approx 2.1213)**Case 4: (beta = frac{pi}{4})**The expression becomes:(3sinalpha cosleft(frac{pi}{4}right) + cosalpha sinleft(frac{pi}{4}right))Again, (cosleft(frac{pi}{4}right) = sinleft(frac{pi}{4}right) = frac{sqrt{2}}{2}), so:(3sinalpha cdot frac{sqrt{2}}{2} + cosalpha cdot frac{sqrt{2}}{2} = frac{3sqrt{2}}{2} sinalpha + frac{sqrt{2}}{2} cosalpha)Factor out (frac{sqrt{2}}{2}):(frac{sqrt{2}}{2} (3sinalpha + cosalpha))Let me denote (h(alpha) = 3sinalpha + cosalpha). To find its maximum over (alpha in [0, frac{pi}{4}]), take the derivative:(h'(alpha) = 3cosalpha - sinalpha)Set derivative equal to zero:(3cosalpha - sinalpha = 0)(3cosalpha = sinalpha)Divide both sides by (cosalpha) (assuming (cosalpha neq 0)):(3 = tanalpha)So, (alpha = arctan(3)). Compute (arctan(3)): approximately 1.2490 radians, which is about 71.56 degrees. But our domain is (alpha in [0, frac{pi}{4}]), which is approximately [0, 0.7854]. So, (arctan(3)) is outside the domain. Therefore, the maximum occurs at one of the endpoints.Evaluate (h(alpha)) at (alpha = 0) and (alpha = frac{pi}{4}):At (alpha = 0):(h(0) = 3sin(0) + cos(0) = 0 + 1 = 1)At (alpha = frac{pi}{4}):(hleft(frac{pi}{4}right) = 3sinleft(frac{pi}{4}right) + cosleft(frac{pi}{4}right) = 3 cdot frac{sqrt{2}}{2} + frac{sqrt{2}}{2} = frac{3sqrt{2}}{2} + frac{sqrt{2}}{2} = 2sqrt{2} approx 2.8284)Therefore, the maximum of (h(alpha)) is (2sqrt{2}) at (alpha = frac{pi}{4}). Thus, the expression when (beta = frac{pi}{4}) is:(frac{sqrt{2}}{2} times 2sqrt{2} = frac{sqrt{2}}{2} times 2sqrt{2} = (sqrt{2} times sqrt{2}) = 2)So, the maximum in this case is 2.Now, let's summarize the maximum values from each case:- Case 1 ((alpha = 0)): (frac{sqrt{2}}{2} approx 0.707)- Case 2 ((alpha = frac{pi}{4})): (sqrt{5} approx 2.236)- Case 3 ((beta = 0)): (frac{3sqrt{2}}{2} approx 2.121)- Case 4 ((beta = frac{pi}{4})): 2So, the maximum among these is (sqrt{5}), which is approximately 2.236.But wait, let me double-check if there are any other critical points on the boundaries. For example, sometimes when dealing with two-variable functions, the maximum can occur at the corners where both variables are at their boundaries. So, let me check the four corners of the domain:1. (alpha = 0), (beta = 0): expression is (sin(0) + 2sin(0) = 0)2. (alpha = 0), (beta = frac{pi}{4}): expression is (sin(-frac{pi}{4}) + 2sin(frac{pi}{4}) = -frac{sqrt{2}}{2} + 2 cdot frac{sqrt{2}}{2} = -frac{sqrt{2}}{2} + sqrt{2} = frac{sqrt{2}}{2} approx 0.707)3. (alpha = frac{pi}{4}), (beta = 0): expression is (sin(frac{pi}{4}) + 2sin(frac{pi}{4}) = frac{sqrt{2}}{2} + 2 cdot frac{sqrt{2}}{2} = frac{sqrt{2}}{2} + sqrt{2} = frac{3sqrt{2}}{2} approx 2.121)4. (alpha = frac{pi}{4}), (beta = frac{pi}{4}): expression is (sin(0) + 2sin(frac{pi}{2}) = 0 + 2 cdot 1 = 2)So, the maximum at the corners is (frac{3sqrt{2}}{2} approx 2.121), which is less than (sqrt{5} approx 2.236). So, indeed, the maximum is (sqrt{5}).But wait, let me just make sure that when (alpha = frac{pi}{4}) and (beta = arctanleft(frac{1}{3}right)), both are within the domain. Since (arctanleft(frac{1}{3}right) approx 0.32175 < frac{pi}{4} approx 0.7854), yes, it is within the domain.Therefore, the maximum value of the expression is (sqrt{5}).But just to be thorough, let me compute the original expression at (alpha = frac{pi}{4}) and (beta = arctanleft(frac{1}{3}right)):First, compute (sin(alpha - beta)) and (sin(alpha + beta)):Given (alpha = frac{pi}{4}), (beta = arctanleft(frac{1}{3}right)).Compute (alpha - beta = frac{pi}{4} - arctanleft(frac{1}{3}right)). Let me compute the sine of this angle.But maybe it's easier to compute numerically.Compute (alpha = frac{pi}{4} approx 0.7854) radians.Compute (beta = arctanleft(frac{1}{3}right) approx 0.32175) radians.So, (alpha - beta approx 0.7854 - 0.32175 = 0.46365) radians.Compute (sin(0.46365)): approximately (sin(0.46365) approx 0.4472).Compute (alpha + beta approx 0.7854 + 0.32175 = 1.10715) radians.Compute (sin(1.10715)): approximately (sin(1.10715) approx 0.8944).Therefore, the expression is:(sin(alpha - beta) + 2sin(alpha + beta) approx 0.4472 + 2 times 0.8944 = 0.4472 + 1.7888 = 2.236), which is approximately (sqrt{5} approx 2.236). So, that checks out.Therefore, I can be confident that the maximum value is indeed (sqrt{5}).**Final Answer**The maximum value is (boxed{sqrt{5}}).

answer：First, I will simplify the given expression by expanding and combining like terms.The original expression is:(2x² - 2y²) - 3(x²y² + x²) + 3(x²y² + y²)Expanding the terms:2x² - 2y² - 3x²y² - 3x² + 3x²y² + 3y²Combining like terms:(2x² - 3x²) + (-2y² + 3y²) + (-3x²y² + 3x²y²)This simplifies to:-x² + y²Now, substituting the given values x = -1 and y = 2:-(-1)² + (2)² = -1 + 4 = 3The final answer is 3.