Let

$$
S_{n}=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{n(n+1)}
$$

calculate the value of $S_{n}$ when 1, 2, 3, and 4 are substituted for $n$; based on the results obtained, write the general form of $S_{n}$; and finally, prove the correctness of this formula by using the method of induction from $n$ to $n+1$.
$$
S_{1}=\frac{1}{2}, S_{2}=\frac{2}{3}, S_{3}=\frac{3}{4}, S_{4}=\frac{4}{5} \text { and so } S_{n}=\frac{n}{n+1}
$$

To prove the correctness of this formula, add the next term to the sum of the first $n$ terms; then

$$
\begin{gathered}
S_{n+1}=\frac{n}{n+1}+\frac{1}{(n+1)+(n+2)}=\frac{n(n+2)+1}{(n+1)+(n+2)}= \\
\frac{(n+1)^{2}}{(n+1)(n+2)}=\frac{n+1}{n+2}
\end{gathered}
$$

We see that the sum of the first $n+1$ terms is given by a formula of the same form as the sum of the first $n$ terms, and thus the formula $S_{n}=\frac{n}{n+1}$ is correct.

(Denes König, Budapest.)

The problem was also solved by: Appel S., Bayer B., Boros J., Czank K., Freibauer E., Kerekes T., Krausz B., Krisztian Gy., Lukhaub Gy., Lupsa Gy., Oblath R., Perl Gy., Sasvari G., Szollósy J., Vajda Ö., Weisz J.