Problem 4. Solve the equation:

$$
(x+1)^{2}+(x+3)^{2}+(x+5)^{2}+\ldots+(x+2021)^{2}=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-2020)^{2}
$$
Answer: $-0.5$.

Solution. The left and right sides of the equation are respectively $1011 x^{2}+2(1+3+5+\ldots+2021) x+(1^{2}+3^{3}+5^{2}+\ldots+2021^{2})$ and $1011 x^{2}-2(2+4+6+\ldots+2020) x+2^{2}+4^{2}+6^{2}+\ldots+2020^{2}$. The equation is reduced to $2(1+2+3+4+\ldots+2021) x=-1^{2}+2^{2}-3^{2}+4^{2}+\ldots-2019^{2}+2020^{2}-2021^{2}$, from which $(1+2021) \cdot 2021 x=-\frac{3+4039}{2} \cdot 1010-2021^{2}$ and $x=\frac{1010-2021}{2022}=-\frac{1}{2}$.