Task 1. Find all natural numbers $n$ such that the sum

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

is a perfect square.
Solution. For $n=2,3,4,5$ we have:

$$
\begin{aligned}
& 1+1 \cdot 2=3 \\
& 1+1 \cdot 2+1 \cdot 2 \cdot 3=9 \\
& 1+1 \cdot 2+1 \cdot 2 \cdot 3+1 \cdot 2 \cdot 3 \cdot 4=33 \\
& 1+1 \cdot 2+1 \cdot 2 \cdot 3+1 \cdot 2 \cdot 3 \cdot 4+1 \cdot 2 \cdot 3 \cdot 4 \cdot 5=153
\end{aligned}
$$

Every next natural number of the form $1+1 \cdot 2+1 \cdot 2 \cdot 3+\ldots+1 \cdot 2 \cdot \ldots \cdot(n-1) n$ ends in 3 because the number $1 \cdot 2 \cdot \ldots \cdot(n-1) n$ contains 2 and 5 as factors, which means it will end in zero. However, the square of a natural number cannot end in 3, so the solutions to the problem are the numbers 1 and 3.