Solve the following equation, where $n \geq 2$ is a given natural number and $x$ is the unknown.

$$
\sum_{i=0}^{n-2} \frac{1}{(x+i)(x+i+1)}=x(x+1)(x+2) \cdot \ldots \cdot(x+n)+\frac{n-1}{x(x+n-1)}
$$
The equation only makes sense with the assumption that the factors in the denominators on both sides are all different from 0. According to this, \( x \neq -i \) and \( x \neq -(i+1) \) for each \( i=0,1, \ldots, n-2 \), that is, \( x \neq 0, -1, -2, \ldots, -(n-1) \).

Each term on the left side can be written as the difference of two fractions:

\[
\frac{1}{(x+i)(x+i+1)}=\frac{(x+i+1)-(x+i)}{(x+i)(x+i+1)}=\frac{1}{x+i}-\frac{1}{x+i+1}
\]

After this decomposition, the subtrahend of each term cancels out with the minuend of the next term, resulting in 0, so only the first term of the difference with \( i=0 \) and the second term of the difference with \( i=(n-2) \) remain. These two terms combined are equal to the second term on the right side:

\[
\frac{1}{x}-\frac{1}{x+(n-1)}=\frac{n-1}{x(x+n-1)}
\]

Therefore, (1) can be rewritten as:

\[
0=x(x+1)(x+2) \cdot \ldots \cdot(x+n-1)(x+n)
\]

The first \( n \) factors being 0 cannot be considered according to the assumption, so the only solution to the equation is from \( x+n=0 \):

\[
x=-n
\]

This indeed makes both sides of (1) equal.

Margit Forró (Komárom, Czechoslovakia, General Secondary School, 3rd grade)