Let's determine the sum of the following series:

$$
s_{n}=\frac{1}{2 \cdot 3 \cdot 4}+\frac{2}{3 \cdot 4 \cdot 5}+\ldots+\frac{n}{(n+1)(n+2)(n+3)}
$$
Let's replace $k(k=1,2, \ldots, n)$ with $(k+1) - 1$ in every numerator, and rearrange the terms as follows:

$$
\begin{aligned}
s_{n} & =\left(\frac{2}{2 \cdot 3 \cdot 4}+\frac{3}{3 \cdot 4 \cdot 5}+\ldots+\frac{n+1}{(n+1)(n+2)(n+3)}\right)- \\
& -\left(\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots+\frac{1}{(n+1)(n+2)(n+3)}\right)
\end{aligned}
$$

The first bracketed sum can be simplified as follows:

$$
\begin{aligned}
& \frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\ldots+\frac{1}{(n+2)(n+3)}=\frac{4-3}{3 \cdot 4}+\frac{5-4}{4 \cdot 5}+\ldots+ \\
& +\frac{(n+3)-(n+2)}{(n+2)(n+3)}=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots+\left(\frac{1}{n+2}-\frac{1}{n+3}\right)= \\
& =\frac{1}{3}-\frac{1}{n+3}=\frac{n}{3(n+3)}
\end{aligned}
$$

The second sum can also be transformed as follows:

$$
\begin{gathered}
\frac{1}{2}\left[\left(\frac{4-3}{2 \cdot 3 \cdot 4}+\frac{3-2}{2 \cdot 3 \cdot 4}\right)+\left(\frac{5-4}{3 \cdot 4 \cdot 5}+\frac{4-3}{3 \cdot 4 \cdot 5}\right)+\ldots+\right. \\
\left.+\left(\frac{(n+3)-(n+2)}{(n+1)(n+2)(n+3)}+\frac{(n+2)-(n+1)}{(n+1)(n+2)(n+3)}\right)\right]= \\
=\frac{1}{2}\left[\frac{1}{2 \cdot 3}-\frac{1}{3 \cdot 4}+\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}+\ldots+\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right]= \\
=\frac{1}{2}\left[\frac{1}{2 \cdot 3}-\frac{1}{(n+2)(n+3)}\right]=\frac{n^{2}+5 n}{12(n+2)(n+3)}
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
s_{n}=\frac{n}{3(n+3)}-\frac{n^{2}+5 n}{12(n+2)(n+3)}=\frac{4 n(n+2)-\left(n^{2}+5 n\right)}{12(n+2)(n+3)}= \\
=\frac{3 n^{2}+3 n}{12(n+2)(n+3)}=\frac{n(n+1)}{4(n+2)(n+3)}
\end{gathered}
$$

Gyula Beke (Hatvan, Bajza g. IV. o. t.)

Note: The formula

$$
s_{n}=\frac{n(n+1)}{4(n+2)(n+3)}
$$

is correct for $n=1$ (since $s_{1}=\frac{1}{24}$), and its general validity can be easily proven by mathematical induction.