1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=6000$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds.
Answer: 12000001

Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 5997$. We get:

$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$

$$
f(6000)-f(5997)=2 \cdot 5997+3
$$

Adding the equations, we get: $f(6000)-f(0)=2 \cdot(0+3+6+\cdots+5997)+3 \cdot 2000=$ $2 \cdot \frac{5997 \cdot 2000}{2}+6000=12000000$. Then $f(6000)=12000001$.