Monday, March 8, 2021

2001 Dutch Math Olympiad

 

Suppose for all x,yRx,yR we have f(x+y)=f(x)+f(y)+xyf(x+y)=f(x)+f(y)+xy and f(4)=10f(4)=10
Let find the value of f(2001)f(2001)
Solution
We will use this theorem to solve the following problem


For all x,yRx,yR we have f(n+1)=f(n)+f(1)+nf(n+1)=f(n)+f(1)+n then
                                    f(n+1)-f(n)=f(1)+nf(n+1)f(n)=f(1)+n we knew that: f(n)=f(n+1)-f(n)f(n)=f(n+1)f(n)
Then, 2000n=02000n=0f(n)=f(2001)-f(0)f(n)=f(2001)f(0)
Or f(2001)=f(2001)=2000n=02000n=0f(n+f(1))f(n+f(1))+f(0)+f(0)  ()()
 As we had: f(x+y)=f(x)+f(y)+xyf(x+y)=f(x)+f(y)+xy when x=y=2x=y=2 then, f(4)=2f(2)+4f(4)=2f(2)+4
                    f(4)=10f(4)=10f(2)=3f(2)=3
We play the same role to find f(1)f(1)
When x=y=1x=y=1 then, f(2)=2f(1)+1f(2)=2f(1)+1f(1)=1f(1)=1
As the same play, we will get f(0)=0f(0)=0
From the equation of ()() , therefore: f(2001)=f(2001)=2000n=02000n=0(n+1)(n+1)+f(0)+f(0)
                    f(2001)=f(2001)=2000n=0(n+1)=2000(2001)2
    
Solution by: Thin Sokkean

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