272. We will prove by induction that $\sum_{j=0}^{k} x_{j}=\frac{1}{(n-k)}, k=$ $0,1, \ldots, n-1$. For $k=0$, the given equality is trivially satisfied by definition. Suppose now that $\sum_{j=0}^{k} x_{j}=\frac{1}{(n-k)}$ for some $k, 0 \leqslant k \leqslant n-2$. Then

$$
\sum_{j=0}^{k+1} x_{j}=x_{k+1}+\sum_{j=0}^{k} x_{j}=\sum_{j=0}^{k} \frac{x_{j}}{n-k-1}+\sum_{j=0}^{k} x_{j}
$$

by the definition of $x_{k+1}$. Using the induction hypothesis, we get

$$
\sum_{j=0}^{k+1} x_{j}=\frac{1}{(n-k-1)(n-k)}+\frac{1}{n-k}=\frac{1}{n-k-1}
$$

From this, taking $k=n-1$, we find the desired sum

$$
\sum_{j=0}^{n-1} x_{j}=1
$$

[E. Lengyel, A. M. M., 76, 86 (January 1969).]
272. Calculation of the sum. Let $n$ be a fixed positive number. Set $x_{0}=\frac{1}{n}$ and $x_{j}=\frac{1}{n-1} \sum_{i=0}^{j-1} x_{i}$ for $j=$ $1,2, \ldots, n-1$. Compute the sum $\sum_{j=0}^{n-1} x_{j}$.