1. Determine the sum of all numbers in the table

$$
\begin{array}{ccccc}
1 & 2 & 3 & \ldots & k \\
2 & 3 & 4 & \ldots & k+1 \\
3 & 4 & 5 & \ldots & k+2 \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
k & k+1 & k+2 & \ldots & 2 k-1
\end{array}
$$
Solution. The sum of Baraniot is equal to:

$$
\sum_{j=1}^{k} \sum_{i=j}^{j+k-1} i=\sum_{j=1}^{k} \frac{k}{2}(2 j+k-1)=\frac{k^{2}(k-1)}{2}+k \sum_{j=1}^{k} j=\frac{k^{2}(k-1)}{2}+k \frac{k(k+1)}{2}=k^{3}
$$