2001 Dutch Math Olympiad

 

Suppose that for all \(x,y\in\mathbb{R}\), we have \(f(x+y)=f(x)+f(y)+xy\) and \(f(4)=10\). Find the value of \(f(2001)\).

Solution

We will use this theorem to solve the following problem

Solution

Given \(f(x+y)=f(x)+f(y)+xy\), let \(y=1\), then \(f(n+1)=f(n)+f(1)+n\), so \(f(n+1)-f(n)=f(1)+n\). Since \(\Delta f(n)=f(n+1)-f(n)\), we have

\[ \sum_{n=0}^{2000}\Delta f(n)=f(2001)-f(0), \]

therefore,

\[ f(2001)=\sum_{n=0}^{2000}(f(1)+n)+f(0). \]

From \(f(4)=10\), let \(x=y=2\):

\[ f(4)=2f(2)+4\Rightarrow f(2)=3. \]

Let \(x=y=1\):

\[ f(2)=2f(1)+1\Rightarrow f(1)=1. \]

Let \(x=y=0\):

\[ f(0)=0. \]

Hence,

\[ f(2001)=\sum_{n=0}^{2000}(n+1)=\frac{2001\times2002}{2}=2003001. \]

Therefore, \(\boxed{f(2001)=2003001}\).

Solution by: Thin Sokkean