2. The function $f$ has the following properties:

$$
\begin{aligned}
& f(x+y)=f(x)+f(y)+x y \text { for all real numbers } x, y . \\
& f(4)=10
\end{aligned}
$$

Calculate $f(2001)$.
2. $10=f(4)=f(2+2)=f(2)+f(2)+4=4 f(1)+6$ so $f(1)=1$.

For any number $n$, it holds that: $f(n+1)-f(n)=f(n)+f(1)+n \times 1-f(n)=n+1$.

Thus, $f(2001)=\{f(2001)-f(2000)\}+\{f(2000)-f(1999)\}+\ldots+\{f(2)-f(1)\}+f(1)=$ $2001+2000+\ldots+2+1=2001 \times 2002 / 2=2003001$.