1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=1500$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds.
Answer: 750001

Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute the numbers $0,3,6, \ldots, 1497$ for $x$. We get:

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

$$
f(1500)-f(1497)=2 \cdot 1497+3
$$

Adding the equations, we get: $f(1500)-f(0)=2 \cdot(0+3+6+\cdots+1497)+3 \cdot 500=$ $2 \cdot \frac{1497 \cdot 500}{2}+1500=750000$. Then $f(1500)=750001$.