Let's determine the sum of the following series:

$$
1 \cdot 2 - 2 \cdot 3 + 3 \cdot 4 - 4 \cdot 5 + \ldots \pm n(n+1)
$$

(Two cases, depending on whether $n$ is even or odd.)
(1) If $n$ is an even number, then

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

$\left(2^{\circ}\right)$ If $n$ is an odd number, then the series can be written as:

$$
\begin{gathered}
=2 \cdot(1-3)+4 \cdot(3-5)+\ldots+(n-1(n-2-n)+n(n+1)= \\
=-2 \cdot(2+4+\ldots+n-1)+n(n+1)= \\
=-2 \cdot(2+n-1) \cdot \frac{n-1}{4}+n(n+1)= \\
=-\frac{(n+1)(n-1)}{2}+n(n+1)=\frac{(n+1)^{2}}{2}
\end{gathered}
$$

(Gyula Krampera, Debrecen.)

Number of solutions: 33.