1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=3000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+3 x+2$ holds.
Answer: 6748501

Solution: In the equation $f(x+2)-f(x)=3 x+2$, we will substitute for $x$ the numbers $0, 2, 4, \ldots, 2998$. We get:

$$
\begin{aligned}
& f(2)-f(0)=3 \cdot 0+2 \\
& f(4)-f(2)=3 \cdot 2+2
\end{aligned}
$$

$$
f(3000)-f(2998)=3 \cdot 2998+2
$$

Adding the equations, we get: $f(3000)-f(0)=3 \cdot(0+2+4+\cdots+2998)+2 \cdot 1500=$ $3 \cdot \frac{2998 \cdot 1500}{2}+3000=6748500$. Then $f(3000)=6748501$.