这玩意第一项都不对啊

应该是减2
机翻的：
Suppose there is an item a_x in this column that satisfies a_x = 3n + 2.
(n is a natural number.)
Then a_(x+1) = 3(3n + 2) - 4 = 9n + 2.
That is: If one item meets the condition, then all subsequent items will also meet the condition.
Verification: a_1 = 5 = 3 + 2
Therefore, all the items in this sequence meet the condition. a_n=2+3^n
假设此列中存在一项a_x满足a_x=3n+2
（n为自然数）
则a_(x+1)=3(3n+2)-4=9n+2
即：若一项满足条件则此后各项都满足条件
验证：a_1=5=3+2
故此数列各项都满足条件
a_n=2+3^n

初中学数列是吧

减2或者加1

it is clear that

证明不了每项减一为三的倍数啊，首先第一项5-1就不是3的倍数

文言文是不是得翻译成古英语，来点Thou Mi啥的

The answer is...if you want it,then you have to take it,but you already knew that

一眼钓鱼

一眼钓鱼题，后续各项一减4马上就符合3的倍数了，但是第一项是5，这么dinner的错误只可能是抖音钓鱼题的出题人自己也这么dinner

3n+2
减一是3n+1，不可能是3的倍数
至于写英语作文，写不了半点

首先出错题了，其次讲解解题思路和正式解题是完全不一样的，讲解会偏口语化和配合手势进行模糊性表达，不符合这种正式写作文的场合。

I don't know

老马家的还行，他发现题出错了就自己给改了一下，让运算正常

Dear Peter,
I'm glad to help you solve this problem from The Nine Chapters on the Mathematical Art.
The sequence starts with a_1 = 5 , and each subsequent term is defined by a_{n+1} = 3a_n - 4 . To find the 12th term, we can derive the general formula a_n = 4 \times 3^{n-1} + 1 . Plugging in n = 12 , we get a_{12} = 708589 .
For the sum of the first n terms, we calculate S_n = 2(3^n - 1) + n .
To prove that a_n - 1 is a multiple of 3, notice that a_n - 1 = 4 \times 3^{n-1} , which is clearly divisible by 3.
Look forward to your early reply.
Yours,
Li Hua