Problem 3. Let $n$ be a positive integer. Calculate the following sum:

$$
\frac{3}{1 \cdot 2 \cdot 4 \cdot 5}+\frac{4}{2 \cdot 3 \cdot 5 \cdot 6}+\frac{5}{3 \cdot 4 \cdot 6 \cdot 7}+\ldots+\frac{n+2}{n \cdot(n+1) \cdot(n+3) \cdot(n+4)}
$$
Solution. Let $S$ be the desired sum. Instead of $S$, we will calculate $2S$, which is written as

$$
\begin{gathered}
2 S=\sum_{k=1}^{n} \frac{2 k+4}{k(k+1)(k+3)(k+4)}=\sum_{k=1}^{n}\left(\frac{1}{k(k+3)}-\frac{1}{(k+1)(k+4)}\right) \\
=\left(\frac{1}{4}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{18}\right)+\ldots+\left(\frac{1}{n(n+3)}-\frac{1}{(n+1)(n+4)}\right) \\
=\frac{1}{4}-\frac{1}{(n+1)(n+4)}=\frac{n(n+5)}{4(n+1)(n+4)}
\end{gathered}
$$

From this, it immediately follows that

$$
S=\frac{n(n+5)}{8(n+1)(n+4)}
$$

Comment. We can proceed in the traditional way by considering the partial fraction decomposition. In this way,

$$
\frac{k+2}{k(k+1)(k+3)(k+4)}=\frac{1 / 6}{k}-\frac{1 / 6}{k+1}-\frac{1 / 6}{k+3}+\frac{1 / 6}{k+4}
$$

and we can proceed in the same way as before, considering the telescoping cancellation.