1. Let $k \in \mathbb{N} \backslash\{1\}$. Calculate the sum of all numbers in the table

$$
\begin{array}{ccccc}
1 & 2 & 3 & \ldots & k \\
2 & 3 & 4 & \ldots & k+1 \\
3 & 4 & 5 & \ldots & k+2 . \\
. & . & . & . & . \\
k & k+1 & k+2 & \ldots & 2 k-1
\end{array}
$$
Solution. Let $j \in\{1,2, \ldots, k\}$. Then the sum of the elements in the $j$-th row is

$$
S_{j}=(j+0)+(j+1)+(j+2)+\ldots+(j+k-1)=k j+\frac{(k-1) k}{2}
$$

Thus, the desired sum is

$$
\begin{aligned}
S_{1}+S_{2}+\ldots+S_{k} & =k \cdot 1+\frac{(k-1) k}{2}+k \cdot 2+\frac{(k-1) k}{2}+k \cdot 3+\frac{(k-1) k}{2}+\ldots+k \cdot k+\frac{(k-1) k}{2} \\
& =k \frac{(k-1) k}{2}+k(1+2+\ldots+k)=k \frac{(k-1) k}{2}+k \frac{(k+1) k}{2} \\
& =\frac{k^{2}}{2}(k-1+k+1)=k^{3}
\end{aligned}
$$