# $x_{n+1}=\sum_{k=1}^{n-1}a_{(n,k)}x_k$, $S=\{x_1,x_2,...\}$ is finite then $0 \in S$

Let $$(R,+,\cdot)$$ be a ring. Let $$M=\{x_1,x_2,...,x_m\}$$ a set of elements from $$R$$.

Consider the following numbers:

$$x_{n+1}=\sum_{k=1}^{n-1}\binom{n}{k}x_k$$ for every $$n\ge m$$

Conjecture:

Prove that if $$S=\{x_1,x_2,...\}$$ is finite then $$0 \in S$$

I was wondering if this is true.

If it is true then:

If it is so, I might be curious in proving the same result but instead the condition

$$x_{n+1}=\sum_{k=1}^{n-1}\binom{n}{k}x_k$$ for every $$n\ge m$$

to have

$$x_{n+1}=\sum_{k=1}^{n-1}a_{(n,k)}x_k$$ for every $$n\ge m$$

where $$a{(n,k)}$$ are some arbitrary elements from $$R$$.

I would highly appreciate if everyone would contribute with something no matter how insignificant so as to find an answer to this conjecture.

• Is the ring assumed to be commutative? May 14, 2022 at 21:51
• No, it isn't commutative May 15, 2022 at 11:16
• Can you please say from where you have got these type of questions? May 16, 2022 at 4:23