1. From the odd natural numbers, we form the sets

$$
A_{1}=\{1\}, \quad A_{2}=\{3,5\}, \quad A_{3}=\{7,9,11\}, \quad A_{4}=\{13,15,17,19\}, \ldots
$$

Calculate the sum of the numbers in the set $A_{n}$.
Solution. Let's find the first number in the set $A_{n}$. The sets $A_{1}, A_{2}, \ldots, A_{n-1}$ are pairwise disjoint, so the number of elements in their union is

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

Thus, the last number in the set $A_{n-1}$ is $n(n-1)-1$. Therefore, the first number in the set $A_{n}$ will be $n(n-1)+1$, and the last number will be $n(n+1)-1$.

The numbers in the set $A_{n}$ form an arithmetic progression with the first term $a_{1}=n(n-1)+1$, the last term $a_{n}=n(n+1)-1$, and the common difference $d=2$. Therefore, their sum will be:

$$
S_{n}=\frac{n}{2}\left(n(n-1)+1+(n(n+1)-1)\right)=n^{3}
$$