-2才对，抖音上这些东西有几个真的？

初中咋可能要150字，高考才是150吧

It is easy to get that An=3^n+2.

有一句肯定用得上，its easy to notice that……

Go fxxk yourself，write your mxxxter

造谣一张图

rt是什么

数学题，那就通篇clearly，obviously就行了

让ai翻译的：The obvious result becomes easily evident, following the previous example clearly. The answer is left as an exercise, and you can prove it on your own without difficulty.
Conversely, by the same reasoning, the conclusion naturally holds. The process is omitted, Q.E.D., as it can be inferred from the above that the proof is complete.

I already have a good idea, but the paper is so thin that I can't show it here.

豆包还是可以的
Dear Peter,
I'm glad to help you solve the problem from The Nine Chapters on the Mathematical Art. Let's break it down step by step. The first term of the sequence is 5, and each subsequent term is 3 times the previous one minus 4.
To find the 12th term: we can derive the recurrence relation a_n = 3a_{n-1} - 4 with a_1=5. By transforming it, we get a_n - 1 = 3(a_{n-1} - 1), so \{a_n - 1\} is a geometric sequence with first term 4 and common ratio 3. Thus, a_n = 4\times3^{n-1} + 1, and a_{12} = 4\times3^{11} + 1 = 708589.
The sum of the first n terms S_n = \sum_{k=1}^n (4\times3^{k-1} + 1) = 2(3^n - 1) + n. Also, since a_n - 1 = 4\times3^{n-1}, it’s obviously a multiple of 3, which proves the statement.
If you have more questions about ancient Chinese math, feel free to ask me!
Look forward to your early reply.
Yours,
LIHUA

九章算术里出现了n也是挺有意思的

最后一句是什么意思？5-1也不是3的倍数啊。

1 : an+1=an*3-4， a1=5，问a12
2: 求Sn
3: 求bn=an-1可被3整除