1A. Let $n$ be a natural number and let $x_{0}=\frac{1}{n}, x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\cdots+x_{k-1}\right)$. Calculate the sum $x_{0}+x_{1}+\cdots+x_{n-1}$.
Solution. Applying the above formula for $x_{k}$, we obtain $x_{1}=\frac{1}{n(n-1)}$, $x_{2}=\frac{1}{(n-1)(n-2)}$. We will prove by mathematical induction that

$$
x_{k}=\frac{1}{(n-k)(n-k+1)}
$$

Let the statement hold for $i<k$. Then

$$
\begin{aligned}
x_{k} & =\frac{1}{n-k}\left(\frac{1}{n}+\frac{1}{n(n-1)}+\frac{1}{(n-1)(n-2)}+\ldots+\frac{1}{(n-k+2)(n-k+1)}\right) \\
& =\frac{1}{n-k}\left(\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n-2}-\frac{1}{n-1}+\ldots+\frac{1}{n-k+1}-\frac{1}{n-k+2}\right) \\
& =\frac{1}{n-k} \cdot \frac{1}{n-k+1}=\frac{1}{(n-k)(n-k+1)}
\end{aligned}
$$

Accordingly,

$$
x_{0}+x_{1}+x_{2}+\cdots+x_{n-1}=\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n-2}-\frac{1}{n-1}+\cdots+\frac{1}{2}-\frac{1}{3}+1-\frac{1}{2}=1
$$