10,11 |

The numbers $1,2, \ldots, k^{2}$ are arranged in a square table

| 1 | 2 | $\ldots$ | $k$ |
| :---: | :---: | :---: | :---: |
| $k+1$ | $k+2$ | $\ldots$ | $2 k$ |
| $\ldots$ | $\ldots$ | $\ldots$ | $\ldots$ |
| $(k-1) k+1$ | $\ldots$ | $\ldots$ | $k^{2}$ |

An arbitrary number is written down, after which the row and column containing this number are crossed out from the table. The same is done with the remaining table of $(k-1)^{2}$ numbers, and so on, $k$ times. Find the sum of the written numbers.
Let's write the given table as

| $k \cdot 0+1$ | $k \cdot 0+2$ | $\ldots$ | $k \cdot 0+k$ |
| :---: | :---: | :---: | :---: |
| $k \cdot 1+1$ | $k \cdot 1+2$ | $\ldots$, | $k \cdot 1+k$ |
| $\ldots$ | $\ldots$ | $\ldots$ | $\ldots$ |
| $(k-1) k+1$ | $(k-1) k+2$ | $\ldots$ | $(k-1) k+k$ |

Each number in the table is represented in the form $k a+b$, where $0 \leq a \leq k-1$ and $1 \leq b \leq k$. We will separately sum the terms $k a$ and the terms $b$. From each row, exactly one number is selected, so there will be terms $k a$ for each $a=0,1, \ldots, k-1$. From each column, exactly one number is selected, so there will be terms $b$ for each $b=1,2, \ldots, k$. Thus, the desired sum is

$\left.k(0+1+2+\ldots+(k-1))+(1+2+\ldots+k)=1 / 2 k(k(k-1)+(k+1))=1 / 2 k\left(k^{2}+1\right)\right)$.

## Answer

$\left.1 / 2 k\left(k^{2}+1\right)\right)$.