6. (20 points) Calculate the value of the expression:

$$
1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots+2019 \cdot 2020 \cdot(2019+2020)
$$
Solution. We will prove by mathematical induction for natural n that:

$$
\begin{gathered}
-0 \cdot 1 \cdot(0+1)+1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots- \\
-(2 n-2) \cdot(2 n-1) \cdot(4 n-3)+(2 n-1) \cdot 2 n \cdot(4 n-1)= \\
=(2 n-1) \cdot 2 n \cdot(2 n+1) .
\end{gathered}
$$

Base: \( n=1 \Rightarrow -0 \cdot 1 \cdot(0+1)+1 \cdot 2 \cdot(1+2)=1 \cdot 2 \cdot 3 \) - true.

Assume it is true for \( n=k \), then for \( n=k+1 \):

$$
\begin{gathered}
-0 \cdot 1 \cdot(0+1)+1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots \\
\ldots+(2 k-1) \cdot 2 k \cdot(4 k-1)- \\
=(2 k-1) \cdot 2 k \cdot(2 k+1)+(2 k+1) \cdot((2 k+2 k) \cdot(4 k+3)-2 k \cdot(4 k+1))= \\
=(2 k+1) \cdot\left(4 k^{2}-2 k+8 k^{2}+14 k+6-8 k^{2}-2 k\right)=(2 k+1) \cdot\left(4 k^{2}+10 k+6\right) \\
=\quad=(2 k+1) \cdot(2 k+2) \cdot(2 k+3), \text { the statement is proved. }
\end{gathered}
$$

Therefore,

\( 1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots+2019 \cdot 2020 \cdot(2019+2020)=2019 \cdot 2020 \cdot \) \( 2021=2020^{3}-2020=8000 \cdot 101^{3}-2020=8000 \cdot 1030301-2020=8242408000-2020= \) 8242405980.

Answer: 8242405980.

| Criterion Content | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 20 |
| There is an arithmetic error or the solution is insufficiently justified in an otherwise correct and justified solution. | 14 |
| The beginning of a possible solution is provided, some intermediate results are obtained (but the formula for \( (2 n-1) \cdot 2 n \cdot(2 n+1) \) is not proved), further solution is incorrect or missing. | 6 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |

## Solution for Variant 7