Problem 1752

[Proof] Take the arithmetic sequence $\left\{a_{n} \mid n \in N\right\}$, where
$a_{1}=x_{1}, a_{2}=x_{2}$, and the common difference $d=x_{2}-x_{1}$.

Notice that
$$\begin{aligned} \sum_{p=0}^{n}(-1)^{p} C_{n}^{p} a_{p+1} & =\sum_{p=0}^{n}(-1)^{p} C_{n}^{p}\left(a_{1}+p d\right) \\ & =a_{1} \sum_{p=0}^{n}(-1)^{p} C_{n}^{p}+d \sum_{p=0}^{n}(-1)^{p} p C_{n}^{p} \end{aligned}$$

And
$$p C_{n}^{p}=p \cdot \frac{n!}{p!(n-p)!}=\frac{n \cdot(n-1)!}{(p-1)!(n-p)!}=n C_{n-1}^{p-1},$$
$$\begin{array}{l} 0=(1-1)^{n}=\sum_{p=0}^{n}(-1)^{p} C_{n}^{p}, \\ 0=(1-1)^{n-1}=\sum_{q=0}^{n-1}(-1)^{q} C_{n-1}^{q}=\sum_{p=0}^{n}(-1)^{p-1} C_{n-1}^{p-1}, \\ \text { (let } q=p-1 \text { ) } \end{array}$$

We have
$$\begin{aligned} \sum_{p=0}^{n}(-1)^{p} C_{n}^{p} a_{n+1} & =d \sum_{p=1}^{n}(-1)^{p} n C_{n-1}^{p-1} \\ & =-d n \sum_{p=1}^{n}(-1)^{p-1} C_{n-1}^{p-1}=0 \end{aligned}$$

Given $x_{1}=a_{1}, x_{2}=a_{2}$.
Assume that we have proved $x_{1}=a_{1}, x_{2}=a_{2}, \cdots, x_{k}=a_{k}$, where $k$ is some number not less than 2.

In the given equation, let $n=k$, we get
$$x_{1}-C_{k}^{1} x_{2}+C_{k}^{2} x_{3}-\cdots+(-1)^{k-1} C_{k}^{k-1} x_{k}+(-1)^{k} C_{k}^{k} x_{k+1}=0,$$

In the equation we have proved
$$\sum_{p=0}^{n}(-1)^{\rho} C_{n}^{\rho} a_{n+1}=0$$

let $n=k$, we get
$$a_{1}-C_{k}^{1} a_{2}+C_{k}^{2} a_{3}-\cdots+(-1)^{k-1} C_{k}^{k-1} a_{k}+(-1)^{k} C_{k}^{k} a_{k+1}=0,$$
(1) - (2), and by the induction hypothesis, we get
$$(-1)^{k} C_{k}^{k}\left(x_{k+1}-a_{k+1}\right)=0$$

Thus, $x_{k+1}=a_{k+1}$.
By the principle of mathematical induction, $x_{n}=a_{n}, n \in N$.
In other words, the sequence $\left\{x_{n} \mid n \in N\right\}$ given in the problem is an arithmetic sequence.
Next, we prove a strengthened proposition: If $\left\{x_{n} \mid n \in N\right\}$ is an arithmetic sequence, then
$$\sum_{p=0}^{n}(-1)^{p} C_{n}^{p} x_{p+1}^{k}=0$$

where $k \in N, n \geqslant k+1$.
Clearly, from the above discussion, the strengthened proposition holds for $k=1$ for all $n \geqslant k+1$.

Assume that for some positive integer $k$, we have already proved
$$\sum_{p=0}^{n}(-1)^{p} C_{m}^{p} x_{p+1}^{k}=0$$

for all $n \geqslant k+1$.
Then, for $n \geqslant(k+1)+1$, we have
$$\begin{aligned} & \sum_{p=0}^{n}(-1)^{p} C_{n}^{p} x_{p+1}^{k+1}=\sum_{p=0}^{n}(-1)^{p} C_{n}^{p} \cdot x_{p+1}^{k} \cdot x_{p+1} \\ = & \sum_{p=0}^{n}(-1)^{p} C_{n}^{p} x_{p+1}^{k} \cdot\left(x_{1}+p d\right) \\ = & x_{1} \sum_{p=0}^{n}(-1)^{p} C_{n}^{p} x_{p+1}^{k}+d \sum_{p=0}^{n}(-1)^{p} \cdot p \cdot C_{n}^{p} \cdot x_{p+1}^{k} \\ = & n d \sum_{p=1}^{n}(-1)^{p} C_{n-1}^{p-1} x_{p+1}^{k} \\ = & -n d \sum_{q=0}^{n-1}(-1)^{q} C_{n-1}^{p} y_{q+1}^{k} . \end{aligned}$$
(let $q=p-1, y_{q+1}=x_{p+1}$)
Since $\left\{x_{n} \mid n \in N\right\}$ is an arithmetic sequence, $\left\{y_{n} \mid n \in N\right\}$ is also an arithmetic sequence. Thus, by the induction hypothesis, we have
$$\sum_{y=0}^{n-1}(-1)^{q} C_{n-1}^{y} y_{q+1}^{k}=0$$

Therefore,
$$\sum_{p=0}^{n}(-1)^{p} C_{n}^{p} x_{p+1}^{k+1}=0 .$$

By the principle of mathematical induction, the strengthened proposition is proved, and thus the original proposition is proved.