Answer: 4074341.

Solution. Note that for any $k$ the equality $k^{2}-(k+1)^{2}-(k+2)^{2}+(k+3)^{2}=4$ holds. Therefore, the entire sum is equal to $1+504 \cdot 4+2018^{2}=4074341$.

| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| $+/-$ | 16 | Correct solution with an arithmetic error |
| $-/+$ | 8 | Correct idea of the solution with completely incorrect calculation and answer |
| - | 0 | Solution is completely incorrect/only answer |
Problem 9-1. Calculate the sum $1^{2}+2^{2}-3^{2}-4^{2}+5^{2}+6^{2}-7^{2}-8^{2}+9^{2}+10^{2}-\ldots+2017^{2}+2018^{2}$.