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 $n\ge m$


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.

  • $\begingroup$ Is the ring assumed to be commutative? $\endgroup$
    – abacaba
    May 14, 2022 at 21:51
  • $\begingroup$ No, it isn't commutative $\endgroup$
    – shangq_tou
    May 15, 2022 at 11:16
  • 1
    $\begingroup$ Can you please say from where you have got these type of questions? $\endgroup$
    – Alexander
    May 16, 2022 at 4:23


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