Question

函数f(x)对于任意实数x,y,都满足f(x+y^2)=f(x)+2f^2(y),且f(1)≠0,则f(1998)=?

Answer 1

令x=0,y=0代入已知,f(0)=f(0)+2f^2(0)∴f(0)=0

x=0,y=1,f(1)=f(0)+2f^2(1),∵ f(1)≠0)∴f(1)=1/2

x=x,y=1,f(x+1)=f(x)+2f^2(1)=f(x)+1/2

∴an=f(n)为首相为f(1),共差为1/2的等差数列

f(n)=an=1/2+（n-1）/2=n/2

∴f(1998)=999