182. Grouped Numbers. Consecutive odd 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.[^10]
182. The $n$-th group contains $n$ integers; therefore, the number of integers in all groups from the 1st to the $n$-th inclusive is $\frac{n(n+1)}{2}$, and the number of integers in all groups up to the $(n-1)$-th inclusive is $\frac{(n-1) n}{2}$. These two sets form arithmetic progressions with a common difference of 2. Therefore, the sum of all numbers in the $n$-th group is

$$
\begin{array}{r}
\frac{1}{2} \cdot \frac{n(n+1)}{2}\left\{2+\left[\frac{n(n+1)}{2}-1\right] 2\right\}- \\
-\frac{1}{2} \cdot \frac{(n-1) n}{2}\left\{2+\left[\frac{(n-1) n}{2}-1\right] 2\right\}= \\
=n^{2} \frac{(n+1)^{2}-(n-1)^{2}}{4}=n^{3}
\end{array}
$$