Question

数列题!f(x,y)对所有实数x,y都满足：f(0,y)=y+1,f(x+1,0)=f(x,1),f(x+1,y+1)=

Answer 1

（1）f(0,y)=y+1,f(x+1,y+1)=f[x,f(x+1,y)],

令x=0,y=n-1得

f(1,n)=f[0,f(1,n-1)]=f(1,n-1)+1,

∴f(1,n)-f(1,n-1)=1,为常数.

（2）f(1,0)=f(0,1)=2,

∴f(1,n)=1+n,

an=f(2,-n)=f[1,f(2,-n-1)]=1+f(2,-n-1),

f(2,0）=f(1,1)=2,

∴an=2-n,

∴{an}是等差数列,

∴a1+a2+...+an=n(a1+an)/2=n(1+2-n)/2=n(3-n)/2.