[Bug] LaTeX from DeepSeek R1 causes frontend errors

[fred913]

📦 Environment

Docker

📌 Version

v1.51.1

💻 Operating System

Windows

🌐 Browser

Edge

🐛 Bug Description

When talking about a math problem it generates response, and then the error page shows.

Response text:

好，我现在得解决这个关于数列的问题。题目给定了一个正向数列{aₙ}，首项a₁=4，递推式是aₙ² = a_{n+1} + 3n aₙ -3。任务分成两部分，首先是根据前几项求出a₂和a₃，再根据这三项的规律来猜出通项公式；第二就是用数学归纳法证明这个猜想正确的。让我仔细分析一下，一步步解决这个问题。

第一部分，求a₂和a₃，然后再猜通项公式。首先，我应该根据递推式来计算a₂的值。注意，这里给出的递推式是aₙ的平方等于后面的项a_{n+1}加3n乘aₙ减3。那么，对于n=1的情况，即当用a₁来计算a₂的时候，我们可以代入到递推式中。让n=1：

a₁的平方等于a₂加上31a₁减去3。现在已知a₁=4，代入进去，4² = a₂ + 314 -3。计算左边是16，右边是a₂ + 12 -3，也就是a₂ +9。所以方程就是16 = a₂ +9，解得a₂ = 16 -9 =7。对吗？得核对一下运算是不是哪里错了。

结果是a₂=7，因为1)带入n=1，式子变成 a₁² =a₂ +31a₁ -3 →4²= a₂ +3*4 -3 →16= a₂ +12 -3 →16=a₂+9 →a₂=7。对吧？

接下来求a₃的话，需要使用a₂的表达式，然后带入n=2时的递推式，同样的方式解出a₃。这时候式子应是a₂² =a₃ +32a₂ -3，根据已知，a₂=7，则带入得：7² = a₃ +327 -3 →49= a₃ +42 -3 →也就是49 =a₃ +39 →解得a₃=49-39=10。所以a₃=10。

现在前三个项是a₁=4，a₂=7，a₃=10。现在要看这三个数的规律，猜想通项公式。他们之间可能隐藏的规律可能是线性的？即相邻两项的差，例如7-4=3，10-7=3，这样下一个可能是a₄=13？或者是否存在某种数列类型。

不过根据题目给出的递推式比较特别，不是线性的递归，而是带有平方项的，是否可能通项公式为一次函数形式，例如 aₙ = 3n +1？

看一下是否是这种可能的，比如当n=1时，31+1=4，确实符合a₁=4；当n=2时，32+1=7，符合a₂=7；当n=3时，3*3+1=10，也符合a₃=10。看起来像是aₙ=3n +1，那么这个猜想似乎合理？或许呢？

好的，接下来是证明部分，需要用数学归纳法来证明这个猜想。

首先，数学归纳法需要两步：基础步骤和归纳步骤。基础步骤要证明当n=1时命题成立，即a₁=4=3*1+1=4，成立。这是正确的。

那么归纳假设设为当n=k时，a_k=3k+1，这假设成立，进而需要证明n=k+1时a_{k+1}=3(k+1)+1=3k+4是否成立。

然而要注意原题使用的是递推式a_n² = a_{n+1} + 3n a_n -3，其中原式中n的范围从1开始到所有自然数。所以，应采用归纳推导出当假设a_k=3k+1成立时，证明a_{k+1}=3(k+1)+1=3k+4.

这个过程可能需要将归纳假设代入递推式进行计算。

但是需要注意这一点不一定直接可行，因为对于给定的递推式，可能需要从n到n+1时的依赖关系来导出后的项是否符合。这个过程需要详细推导是否正确。或者说： 假设对于任意的n=k，a_k=3k+1，那么递推式a_k² =a_{k+1} +3k*a_k -3里面的a_{k+1}会等于什么呢？

带入a_k=3k+1，左边a_k²=(3k+1)^2=9k² +6k +1；右边则是a_{k+1} +3k*(3k+1) -3.

根据等式：9k²+6k+1 = a_{k+1} +3k*(3k+1) -3.

右边的计算: 3k*(3k+1)是9k² +3k，整个右边的表达式为 a_{k+1} +9k² +3k -3。此时左边等于右边，所以方程是：9k² +6k +1 = a_{k+1} +9k² +3k -3.

简化的话，可以先对两边约去相同的项。左边的9k²=右边的9k²，剩下的左边是6k +1，右边对剩下的项是 a_{k+1} +3k -3。

这样就得等式：6k +1 = a_{k+1} +3k -3.

把左边的6k +1 变成右边的话，则a_{k+1} +3k -3=6k +1 →解，移项得a_{k+1}=6k +1 -3k +3=3k +4。因此，这个过程对吗？

这样可以得出 a_{k+1}=3k+4=3(k+1)+1，即确实满足 a_{k+1}=3(k+1)+1。这说明如果n=k时，a_k=3k+1成立，则n=k+1时，a_{k+1}=3(k+1)+1成立。这说明通过数学归纳法，对于所有自然数n≥1，该公式成立。

那看来这个猜想是对的，所以问题得以解决。

但有一丝问题：是否有可能前几项符合某个规律但后面不符，这题设置显然是通过递推式给出的，且替换正确的情况下，这里归纳法是应用正确的？是的，因为在递推式之下，每个项都依赖前一项，因此数学归纳法适合这种情况。

总之，第一步求出a₂=7和a₃=10，前三项呈现出线性规律：aₙ=3n+1。通过数学归纳法验证，当假设n=k成立时，n=k+1也成立，所以整个通项公式成立。

现在需要系统化地把这个过程整理下来。

答案

1. 代入递推式求得a₂=7，a₃=10。猜想通项公式为$\boxed{a_n = 3n + 1}$。

2. 数学归纳法证明过程如下：

   • 基例：当n=1时，a₁=4=3×1+1，成立。
   • 假设：当n=k时，a_k=3k+1成立。
   • 递推：将a_k代入递推式得：
     [
     (3k+1)^2 = a_{k+1} + 3k(3k+1) - 3
     ]
     计算得a_{k+1}=3(k+1)+1，即通项公式对n=k+1成立。
   • 结论：猜想成立，通项公式为$\boxed{a_n = 3n + 1}$。

知识点回顾

1. 递推数列求解：通过给定的递推关系式依次计算数列的项。
2. 数学归纳法：一种证明方法，用于验证某命题对所有自然数成立。

---

详细解题步骤

问题1：求 ( a_2 )、( a_3 ) 并猜想通项公式

1. 计算 ( a_2 )
   根据递推式 ( a_n^2 = a_{n+1} + 3n a_n - 3 )，当 ( n=1 ) 时：
   [
   a_1^2 = a_2 + 3 \cdot 1 \cdot a_1 - 3
   ]
   已知 ( a_1 = 4 )，代入：
   [
   4^2 = a_2 + 3 \cdot 1 \cdot 4 - 3 \implies 16 = a_2 + 9 \implies a_2 = 7
   ]

2. 计算 ( a_3 )
   当 ( n=2 ) 时，代入递推式：
   [
   a_2^2 = a_3 + 3 \cdot 2 \cdot a_2 - 3
   ]
   已知 ( a_2 = 7 )，代入：
   [
   7^2 = a_3 + 42 - 3 \implies 49 = a_3 + 39 \implies a_3 = 10
   ]

3. 猜想通项公式
   观察前三项：( a_1 = 4 )，( a_2 = 7 )，( a_3 = 10 )，可见 ( a_n ) 是线性递增的，首次差为 ( 3 )。因此猜想通项公式为：
   [
   a_n = 3n + 1
   ]

---

问题2：用数学归纳法证明猜想

1. 基例验证 (n=1)
   当 ( n=1 ) 时，( a_1 = 3 \cdot 1 + 1 = 4 )，与已知相符。

2. 归纳假设
   假设当 ( n = k ) 时，( a_k = 3k + 1 ) 成立。

3. 归纳步骤 (证明 ( n = k+1 ))
   根据递推式，将 ( a_k = 3k + 1 ) 代入：
   [
   (3k + 1)^2 = a_{k+1} + 3k(3k + 1) - 3
   ]
   展开并化简左边：
   [
   9k^2 + 6k + 1 = a_{k+1} + 9k^2 + 3k - 3
   ]
   消去 ( 9k^2 )，并整理：
   [
   6k + 1 = a_{k+1} + 3k - 3 \implies a_{k+1} = 3k + 4
   ]
   这正是 ( n = k+1 ) 时的猜想通项：
   [
   a_{k+1} = 3(k + 1) + 1
   ]

4. 结论
   由数学归纳法可知，通项公式对所有自然数 ( n ) 成立。

---

最终答案

1. 计算结果与猜想：
   [
   a_2 = 7,\quad a_3 = 10,\quad \boxed{a_n = 3n + 1}
   ]

2. 数学归纳法验证成功：
   [
   \boxed{a_n = 3n + 1}
   ]

📷 Recurrence Steps

Send the math problem.

设正向数列 $ { a_n } $ 的首项为 $ 4 $ ，满足 $ a^2_n = a_{n+1} + 3na_n - 3 $.

1. 求 $ a_2 $ 和 $ a_3 $，根据前三项的规律猜想该数列的通项公式
2. 用数学归纳法证明你的猜想。

🚦 Expected Behavior

Correctly display the solution of the math problem

📝 Additional Information

Frontend logs are uploaded on PasteBin

[lobehubbot]

👀 @fred913

Thank you for raising an issue. We will investigate into the matter and get back to you as soon as possible.
Please make sure you have given us as much context as possible.
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[lobehubbot]

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---

📦 Environment

Docker

📌 Version

v1.51.1

💻 Operating System

Windows

🌐 Browser

Edge

🐛 Bug Description

When talking about a math problem it generates response, and then the error page shows.

Response text:

"

OK, I have to solve this problem with the sequence now. The question gives a forward sequence {aₙ}, the first term a₁=4, and the recursive formula is aₙ² = a_{n+1} + 3n aₙ -3. The task is divided into two parts. First, we calculate a₂ and a₃ based on the first few terms, and then guess the general formula based on the rules of these three terms; the second is to use mathematical induction to prove that this conjecture is correct. Let me analyze it carefully and solve this problem step by step.
"
Part 1, find a₂ and a₃, and then guess the term formula. First, I should calculate the value of a₂ based on the recursive formula. Note that the recursive formula given here is that the square of aₙ is equal to the following term a_{n+1} plus 3n times aₙ to minus 3. Then, for the case where n=1, that is, when a₁ is used to calculate a₂, we can substitute it into the recursive formula. Let n=1:
"
The square of a₁ is equal to a₂ plus 31a₁ minus 3. Now it is known that a₁=4, substitute it, 4² = a₂ + 314 -3. Calculate that the left is 16 and the right is a₂ + 12 -3, that is, a₂ +9. So the equation is 16 = a₂ +9, and the solution is a₂ = 16 -9 =7. Is it right? You have to check if the operation is wrong.
"
The result is a₂=7, because 1) brings n=1, and the formula becomes a₁² =a₂ +31a₁ -3 →4²= a₂ +34 -3 →16= a₂ +12 -3 → 16=a₂+9 →a₂=7. Right?
"
Next, if you want a₃, you need to use the expression of a₂, and then bring the recursive formula when n=2, and solve a₃ in the same way. At this time, the formula should be a₂² =a₃ +32a₂ -3. According to the known, a₂=7, it will bring it into: 7² = a₃ +327 -3 →49= a₃ +42 -3 → That is, 49 = a₃ +39 →The solution is to a₃=49-39=10. So a₃=10.
"
Now the first three terms are a₁=4, a₂=7, a₃=10. Now we need to look at the rules of these three numbers and guess the general formula. The law that might be hidden between them may be linear? That is, the difference between two adjacent terms, such as 7-4=3, 10-7=3, so the next one may be a₄=13? Or whether there is some sort of sequence type.
"
However, the recursion formula given according to the question is more special. It is not linear recursion, but with square terms. Is it possible that the general term formula is a one-time function form, such as aₙ = 3n +1?
"
Check whether this is possible. For example, when n=1, 31+1=4, it does meet a₁=4; when n=2, 32+1=7, it meets a₂=7; When n=3, 33+1=10 also meets a₃=10. It looks like aₙ=3n +1, so this conjecture seems reasonable? Maybe?
"
OK, next is the proof part, which needs to be proved by mathematical induction.
"
First, mathematical induction requires two steps: basic steps and induction steps. The basic steps should prove that when n=1 the proposition is true, that is, a₁=4=31+1=4, which is true. that's right.
"
Then the induction assumption is set to when n=k, a_k=3k+1, which is true, and it is necessary to prove that when n=k+1 a_{k+1}=3(k+1)+1=3k+4 Whether it is true.
"
However, it should be noted that the original question uses the recursive formula a_n² = a_{n+1} + 3n a_n -3, where the range of n in the original formula starts from 1 to all natural numbers. Therefore, induction should be used to deduce that when the assumption a_k=3k+1 is true, prove that a_{k+1}=3(k+1)+1=3k+4.
"
This process may require the inductive assumption to be substituted into recursive formulas for calculation.
"
However, it is important to note that this is not necessarily feasible, because for a given recursive formula, the dependency from n to n+1 may be required to derive the following terms. This process requires detailed deduction whether it is correct. Or: Assume that for any n=k, a_k=3k+1, then what will the a_{k+1} in the recursive formula a_k² =a_{k+1} +3ka_k -3 equal?
"
Bring in a_k=3k+1, on the left a_k²=(3k+1)^2=9k² +6k +1; on the right a_{k+1} +3k*(3k+1) -3.
"
According to the equation: 9k²+6k+1 = a_{k+1} +3k*(3k+1) -3.
"
Calculation on the right: 3k*(3k+1) is 9k² +3k, and the expression on the whole right is a_{k+1} +9k² +3k -3. At this time, the left side is equal to the right side, so the equation is: 9k² +6k +1 = a_{k+1} +9k² +3k -3.
"
To simplify, you can first make the same items on both sides. The left 9k² = the right 9k², the left left is 6k +1, and the left term on the right is a_{k+1} +3k -3.
"
This gives the equation: 6k +1 = a_{k+1} +3k -3.
"
If the 6k on the left +1 is turned into the right, then a_{k+1} +3k -3=6k +1 → Solution, and the term is moved to a_{k+1}=6k +1 -3k +3=3k + 4. So, is this process right?
"
This results in a_{k+1}=3k+4=3(k+1)+1, that is, a_{k+1}=3(k+1)+1 is indeed satisfied. This means that if a_k=3k+1 is true when n=k+1 is true when a_{k+1}=3(k+1)+1 is true. This shows that through mathematical induction, this formula holds true for all natural numbers n≥1.
"
It seems that this conjecture is correct, so the problem is solved.
"
But there is a little problem: is it possible that the first few items meet a certain rule but the latter does not meet? This question is obviously given through recursive form, and the replacement is correct. Is the induction here the correct application? Yes, because under recursive, each term depends on the previous one, so mathematical induction is suitable for this situation.
"
In short, the first step is to find a₂=7 and a₃=10. The first three terms show a linear law: aₙ=3n+1. Through mathematical induction verification, when the assumption that n=k is true, n=k+1 is also true, so the entire general formula is true.
"
Now we need to systematically sort out this process.
"
"
Answer
"

1. Substitute the recursive formula to find a₂=7, a₃=10. The conjecture formula is $\boxed{a_n = 3n + 1}$.
   "
2. The mathematical induction proof process is as follows:

• Basic Example: When n=1, a₁=4=3×1+1, which is true.
• Suppose: When n=k, a_k=3k+1 holds.
• Recursive: Substitute a_k into the recursive form to get:
  [
  (3k+1)^2 = a_{k+1} + 3k(3k+1) - 3
  ]
  Calculate a_{k+1}=3(k+1)+1, that is, the general formula is valid for n=k+1.
• Conclusion: The conjecture is true, and the general formula is $\boxed{a_n = 3n + 1}$.

"

Review of knowledge points

"

1. Recursive sequence solution: Calculate the terms of the sequence in sequence through the given recursive relationship.
2. Mathematical Induction: A proof method used to verify that a proposition holds for all natural numbers.
   "

---

"

Detailed solution steps

"

Question 1: Find ( a_2 ), ( a_3 ) and guess the general term formula

"

1. Calculation ( a_2 )
   According to the recursive expression ( a_n^2 = a_{n+1} + 3n a_n - 3 ), when ( n=1 ):
   [
   a_1^2 = a_2 + 3 \cdot 1 \cdot a_1 - 3
   ]
   Known ( a_1 = 4 ), substitute:
   [
   4^2 = a_2 + 3 \cdot 1 \cdot 4 - 3 \implies 16 = a_2 + 9 \implies a_2 = 7
   ]
   "
2. Calculation ( a_3 )
   When ( n=2 ), substitute for recursive expression:
   [
   a_2^2 = a_3 + 3 \cdot 2 \cdot a_2 - 3
   ]
   Known ( a_2 = 7 ), substitute:
   [
   7^2 = a_3 + 42 - 3 \implies 49 = a_3 + 39 \implies a_3 = 10
   ]
   "
3. Conjecture formula
   Observe the first three terms: ( a_1 = 4 ), ( a_2 = 7 ), ( a_3 = 10 ), it can be seen that ( a_n ) is linearly incremented, and the first difference is ( 3 ). Therefore, the conjecture formula is:
   [
   a_n = 3n + 1
   ]
   "

---

"

Question 2: Prove conjectures using mathematical induction

"

1. Basic Example Verification (n=1)
   When ( n=1 ), ( a_1 = 3 \cdot 1 + 1 = 4 ), which matches the known.
   "
2. Inductive Hypothesis
   Assume that when ( n = k ), ( a_k = 3k + 1 ) holds.
   "
3. Induction steps (Prove ( n = k+1 ))
   According to the recursive formula, substitute ( a_k = 3k + 1 ) into:
   [
   (3k + 1)^2 = a_{k+1} + 3k(3k + 1) - 3
   ]
   Expand and simplify the left:
   [
   9k^2 + 6k + 1 = a_{k+1} + 9k^2 + 3k - 3
   ]
   Eliminate ( 9k^2 ) and organize:
   [
   6k + 1 = a_{k+1} + 3k - 3 \implies a_{k+1} = 3k + 4
   ]
   This is exactly the conjecture term when ( n = k+1 ):
   [
   a_{k+1} = 3(k+1) + 1
   ]
   "
4. Conclusion
   From mathematical induction, we can see that the general term formula holds for all natural numbers ( n ).
   "

---

"

Final answer

"

1. ** Calculation results and conjectures**:
   [
   a_2 = 7,\quad a_3 = 10,\quad \boxed{a_n = 3n + 1}
   ]
   "
2. Mathematical induction verification was successful:
   [
   \boxed{a_n = 3n + 1}
   ]

📷 Recurrence Steps

Send the math problem.

Suppose the first term of the forward sequence $ { a_n } $ is $ 4 $ , satisfying $ a^2_n = a_{n+1} + 3na_n - 3 $.
"

1. Find $ a_2 $ and $ a_3 $, and guess the general formula of the sequence according to the law of the first three terms
2. Prove your conjecture using mathematical induction.

🚦 Expected Behavior

NO Response

📝 Additional Information

Frontend logs are uploaded on PasteBin

[fred913]

Updated the issue body but the bot doesn't update translations; Except Behabior is 'Correctly show the solution to the math problem'

[lobehubbot]

✅ @fred913

This issue is closed, If you have any questions, you can comment and reply.
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[lobehubbot]

🎉 This issue has been resolved in version 1.53.5 🎉

The release is available on:

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