| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | C | C |
| deepseekr1 | To determine the total distance Alex walked after placing the first 50 stones, we analyze each trip. For the first stone, he walks 3 feet each way (total 6 feet). Each subsequent trip increases by 2 feet each way, making the total per trip increase by 4 feet (2 feet each way × 2). The distances form an arithmetic sequence: - First term (\(a_1\)) = 6 feet (3 out + 3 back) - Common difference (\(d\)) = 4 feet (additional 2 feet each way per stone) - Number of terms (\(n\)) = 50 The \(n^\text{th}\) term (\(a_n\)) is calculated as: \[ a_{50} = a_1 + (n-1)d = 6 + (50-1) \times 4 = 6 + 196 = 202 \text{ feet} \] The sum of the arithmetic series (\(S_n\)) is: \[ S_{50} = \frac{n}{2} \times (a_1 + a_{50}) = \frac{50}{2} \times (6 + 202) = 25 \times 208 = 5200 \text{ feet} \] Thus, the total distance Alex walked is **5200 feet**, corresponding to option **C**. **Answer: C** |
C |
| gpto1 | C | C |
| GPT-4o | To determine the total distance Alex walked, we need to calculate the distance for each trip to place a stone and then sum these distances. 1. For the first stone: - Walk to place the stone: 3 feet - Walk back to the crate: 3 feet - Total for the first stone: \(3 + 3 = 6\) feet 2. For the second stone: - Walk to place the stone: \(3 + 2 = 5\) feet - Walk back to the crate: 5 feet - Total for the second stone: \(5 + 5 = 10\) feet 3. For the third stone: - Walk to place the stone: \(3 + 2 \times 2 = 7\) feet - Walk back to the crate: 7 feet - Total for the third stone: \(7 + 7 = 14\) feet This pattern continues, with the distance increasing by 2 feet each way for each subsequent stone. Therefore, the distance for the \(n\)-th stone is: - Walk to place the stone: \(3 + 2(n-1)\) feet - Walk back to the crate: \(3 + 2(n-1)\) feet - Total for the \(n\)-th stone: \(2 \times (3 + 2(n-1)) = 6 + 4(n-1)\) feet Now we need to sum this distance for the first 50 stones: \[ \text{Total distance} = \sum_{n=1}^{50} (6 + 4(n-1)) \] \[ = \sum_{n=1}^{50} 6 + \sum_{n=1}^{50} 4(n-1) \] \[ = 6 \times 50 + 4 \sum_{n=1}^{50} (n-1) \] \[ = 300 + 4 \times \sum_{n=0}^{49} n \] \[ = 300 + 4 \times \frac{49 \times 50}{2} \] \[ = 300 + 4 \times 1225 \] \[ = 300 + 4900 \] \[ = 5200 \text{ feet} \] Therefore, the total distance Alex walked is: \[ \boxed{C. 5200} \] |
C |
| llama3.1-70B-Instruct | B |
C |
| llama3.1-405B-Instruct | B. 90950 |
C |
| DeepSeek-Coder-V2 | B. 90950 | C |
| Qwen2-72B-Instruct | B. 90950 | C |
| baichuan4 | B | C |
| doubao | For the first stone, Alex walks 3 feet forward and 3 feet back, so that's 6 feet. For the second stone, Alex walks 5 feet forward and 5 feet back, so that's 10 feet. For the third stone, Alex walks 7 feet forward and 7 feet back, so that's 14 feet and so on. This forms an arithmetic sequence with a first term of 6 and a common difference of 4. The sum of an arithmetic sequence can be found using the formula Sn = n/2(2a + (n-1)d). For the first 50 stones, n = 50, a = 6, and d = 4. Plugging these values in, we get: S50 = 50/2(2*6 + (50-1)*4) = 25(12 + 49*4) = 25(12 + 196) = 25*208 = 5200 feet. So the answer is C. |
C |
| minimax | To solve this problem, we need to calculate the total distance walked by Alex for each of the 50 stones and then sum these distances. For the first stone, Alex walks 3 feet to place it and then 3 feet to return, for a total of 6 feet. For the second stone, Alex walks 5 feet (3 feet to place it and 2 feet back), for a total of 10 feet for the second stone. For the third stone, Alex walks 7 feet (3 feet to place it and 4 feet back), for a total of 14 feet for the third stone. This pattern continues, with each round trip being 4 feet longer than the previous one (since Alex walks 2 feet farther each way). The distance walked for each stone forms an arithmetic sequence: - First stone: 6 feet - Second stone: 6 + 4 = 10 feet - Third stone: 10 + 4 = 14 feet - ... - 50th stone: 6 + 4(50 - 1) = 6 + 4(49) = 6 + 196 = 202 feet The sum of an arithmetic series can be calculated using the formula: Sum = n/2 * (first term + last term) where n is the number of terms. In this case, the first term is 6 feet (for the first stone), the last term is 202 feet (for the 50th stone), and there are 50 terms. Sum = 50/2 * (6 + 202) Sum = 25 * 208 Sum = 5200 feet So, after returning to the crate from placing the 50th stone, the total distance Alex walked is 5200 feet. The correct answer is C. 5200. |
C |
| kimi | To solve this problem, we need to calculate the total distance Alex walked for each stone and then sum up all those distances. For the first stone, Alex walks 3 feet out and 3 feet back, which is a total of 6 feet. For each subsequent stone, Alex walks 2 feet farther each way than the previous stone. So for the second stone, Alex walks 5 feet out and 5 feet back, which is a total of 10 feet. This pattern continues, with the distance walked for each stone forming an arithmetic sequence where the first term \(a_1 = 6\) feet and the common difference \(d = 4\) feet (since Alex walks 2 feet farther each way, the total distance increases by 4 feet each time). The total distance walked for the \(n\)-th stone can be calculated using the formula for the sum of an arithmetic sequence: \[ S_n = \frac{n}{2} (2a_1 + (n-1)d) \] For the 50th stone, we have: \[ S_{50} = \frac{50}{2} (2 \times 6 + (50-1) \times 4) \] \[ S_{50} = 25 (12 + 196) \] \[ S_{50} = 25 \times 208 \] \[ S_{50} = 5200 \] So the total distance Alex walked after placing the 50th stone is 5200 feet. Answer: C. 5200 |
C |
| wenxinyiyan | C. 5200 Explanation: This is a problem involving arithmetic sequences. For each stone placed, Alex walks a certain distance to place the stone and then walks the same distance back to the crate. The distances form an arithmetic sequence with a common difference of 4 feet (2 feet more each way). For the first stone, Alex walks 3 feet to place the stone and 3 feet back to the crate, for a total of 6 feet. The distances walked for each stone, in feet, are as follows: * For the $1^\text{st}$ stone: 6 * For the $2^\text{nd}$ stone: 6 + 4 = 10 * For the $3^\text{rd}$ stone: 10 + 4 = 14 ... and so on. This is an arithmetic sequence with the first term $a_1 = 6$ and the common difference $d = 4$. To find the sum of the distances walked for all 50 stones, we use the formula for the sum of an arithmetic sequence: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$where $S_n$ is the sum of the first $n$ terms, $a_1$ is the first term, $d$ is the common difference, and $n$ is the number of terms. Substituting the given values ($a_1 = 6$, $d = 4$, $n = 50$) into the formula, we get: $S_{50} = \frac{50}{2}(2 \times 6 + (50-1) \times 4)$$S_{50} = 25 \times (12 + 196)$$S_{50} = 25 \times 208$$S_{50} = 5200$Therefore, the total distance Alex walked is 5200 feet. |
C |
| zhipu | To determine the total distance Alex walked, let's first understand the pattern of the distances he walks for each stone. - For the 1st stone: Alex walks 3 feet to place the stone and 3 feet back, totaling \(2 \times 3 = 6\) feet. - For the 2nd stone: Alex walks \(3 + 2 = 5\) feet to place the stone and 5 feet back, totaling \(2 \times 5 = 10\) feet. - For the 3rd stone: Alex walks \(5 + 2 = 7\) feet to place the stone and 7 feet back, totaling \(2 \times 7 = 14\) feet. This pattern continues such that for the \(n\)-th stone, Alex walks \(2 \times (3 + 2(n-1))\) feet. Let's denote the distance walked for the \(n\)-th stone as \(d_n\): \[ d_n = 2 \times (3 + 2(n-1)) \] \[ d_n = 2 \times (3 + 2n - 2) \] \[ d_n = 2 \times (2n + 1) \] \[ d_n = 4n + 2 \] To find the total distance Alex walked for the first 50 stones, we sum \(d_n\) from \(n = 1\) to \(n = 50\): \[ \text{Total distance} = \sum_{n=1}^{50} (4n + 2) \] We can split the sum into two parts: \[ \text{Total distance} = \sum_{n=1}^{50} 4n + \sum_{n=1}^{50} 2 \] The first sum is: \[ \sum_{n=1}^{50} 4n = 4 \sum_{n=1}^{50} n \] \[ \sum_{n=1}^{50} n = \frac{50 \times 51}{2} = 1275 \] \[ \sum_{n=1}^{50} 4n = 4 \times 1275 = 5100 \] The second sum is: \[ \sum_{n=1}^{50} 2 = 2 \times 50 = 100 \] Adding these together: \[ \text{Total distance} = 5100 + 100 = 5200 \] Thus, the total distance Alex walked is \(5200\) feet. The correct answer is: C. 5200 |
C |
| gpto1 | C | C |
| deepseekv3 | To calculate the total distance Alex walked, let's break down the problem step by step. 1. **First Stone:** - **Outbound:** 3 feet - **Return:** 3 feet - **Total for first stone:** \(3 + 3 = 6\) feet 2. **Subsequent Stones:** - For each subsequent stone, Alex walks 2 feet farther each way than the previous stone. - **Outbound distance for the \(n^\text{th}\) stone:** \(3 + 2(n - 1)\) feet - **Return distance for the \(n^\text{th}\) stone:** \(3 + 2(n - 1)\) feet - **Total distance for the \(n^\text{th}\) stone:** \(2 \times [3 + 2(n - 1)] = 6 + 4(n - 1)\) feet 3. **Total Distance for 50 Stones:** - The total distance can be calculated using the formula for the sum of an arithmetic series: \[ \text{Total Distance} = \sum_{n=1}^{50} [6 + 4(n - 1)] = 50 \times 6 + 4 \times \sum_{n=1}^{50} (n - 1) \] - Simplifying the sum: \[ \sum_{n=1}^{50} (n - 1) = \sum_{k=0}^{49} k = \frac{49 \times 50}{2} = 1225 \] - Therefore: \[ \text{Total Distance} = 300 + 4 \times 1225 = 300 + 4900 = 5200 \text{ feet} \] **Final Answer:** \(\boxed{C}\) |
C |