3. For the quadratic trinomials $f_{1}(x)=a x^{2}+b x+c_{1}, f_{2}(x)=a x^{2}+b x+c_{2}$, $\ldots, f_{2020}(x)=a x^{2}+b x+c_{2020}$, it is known that each of them has two roots.

Denote by $x_{i}$ one of the roots of $f_{i}(x)$, where $i=1,2, \ldots, 2020$. Find the value

$$
f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\cdots+f_{2020}\left(x_{2019}\right)+f_{1}\left(x_{2020}\right)
$$
Answer: 0.

## Solution:

Since $f_{1}\left(x_{1}\right)=0$, then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$.

Similarly, we can obtain the following equalities:

$$
f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_{2020}\left(x_{2019}\right)=c_{2020}-c_{2019}, f_{1}\left(x_{2020}\right)=c_{1}-c_{2020}
$$

Adding these equalities, we get

$$
f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\cdots+f_{2020}\left(x_{2019}\right)+f_{1}\left(x_{2020}\right)=0
$$