The odd natural numbers are grouped as follows:

$$
\{1\},\{3,5\},\{7,9,11\},\{13,15,17,19\}, \ldots
$$

Find the sum of the numbers in the $n$-th group. Justify your answer.
Solution. It is observed that the sought sum is probably $n^{3}$....

It is easily shown by induction (or directly) that the sum of the first $n$ odd natural numbers is $S_{n}=n^{2}$ $2 p$

In the first $n$ groups, there are $1+2+\ldots+n=\frac{n(n+1)}{2}$ odd numbers, and in the first $n-1$ groups, $\frac{n(n-1)}{2}$ odd numbers.

We deduce that the sought sum is

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