2001 Dutch Math Olympiad

 

Suppose for all `x,y\inR` we have `f(x+y)=f(x)+f(y)+xy` and `f(4)=10`. 

Let find the value of `f(2001)`

Solution

We will use this theorem to solve the following problem

For all `x,y\inR` we have `f(n+1)=f(n)+f(1)+n` then

                                    `f(n+1)-f(n)=f(1)+n` we knew that: `\triangle``f(n)=f(n+1)-f(n)`

Then, `\sum_{n=0}^{2000}``\triangle``f(n)=f(2001)-f(0)`

Or `f(2001)=``\sum_{n=0}^{2000}``f(n+f(1))``+f(0)`  `(*)`

 As we had: `f(x+y)=f(x)+f(y)+xy` when `x=y=2` then, `f(4)=2f(2)+4`

                    `f(4)=10``\rightarrow``f(2)=3`

We play the same role to find `f(1)`

When `x=y=1` then, `f(2)=2f(1)+1``\rightarrow``f(1)=1`

As the same play, we will get `f(0)=0`

From the equation of `(*)` , therefore: `f(2001)=``\sum_{n=0}^{2000}``(n+1)``+f(0)`

                    `f(2001)=``\sum_{n=0}^{2000}``(n+1)``=(2000(2001))/2`

    

Solution by: Thin Sokkean