1. Let $n$ be a natural number, let $x_{0}=\frac{1}{n}$ and for each $k=1,2, \ldots ., n-1$ let

$$
x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right)
$$

Calculate the sum $S_{k}=x_{0}+x_{1}+\ldots+x_{k}$.
Solution. For $k=0$ and $k=1$ we have

$$
\begin{aligned}
& S_{0}=x_{0}=\frac{1}{n} \\
& S_{1}=x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} x_{0}=\frac{1}{n}+\frac{1}{n-1} \frac{1}{n}=\frac{1}{n-1}
\end{aligned}
$$

We can assume that $S_{k}=\frac{1}{n-k}$. For $k=0$ and $k=1$ this is correct. Let's assume that the given formula is correct for $k \leq s$. Then for $k=s+1$ we have

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

By the principle of mathematical induction, we obtain that for every $k$ it holds that $S_{k}=\frac{1}{n-k}$.