f(x)=ax^2+bx+c
f(0)=0+0+c=0
所以f(x)=ax^2+bx
f(x+1)=a(x+1)^2+b(x+1)
=ax^2+(2a+b)x++(a+b)
=f(x)+x+1
=ax^2+(b+1)x+1
对应项系数相等
所以2a+b=b+1
a+b=1
所以a=1/2,b=1/2
f(x)=(1/2)x^2+(1/2)x