Question

高中函数数列综合

Answer 1

1）
f(2)=f(2*1)=f(2)+f(1)=1
=>
f(1)=0
=>
f(1)=f(2*0.5)=f(2)+f(0.5)=0
=>
f(0.5)=-f(2)=-1

2)
设k>1，则kx>x且f(k)>0
=>
f(k*x)=f(k)+f(x)>f(x)
=>
递增

3)
f(Sn)=f(An)+f(An+1)-1
=>
f(Sn)=f(An)+f(An+1)-f(2)
=>
f(Sn)-f(An)=f(An+1)-f(2)
=>
f(Sn/An)=f((An+1)/2)
有递增性质知：
Sn/An=(An+1)/2
=>
Sn=An*(An+1)/2
=>
S(n-1)=A(n-1)*[A(n-1)+1]/2
=>
Sn-S(n-1)={An*(An+1)/2}-{A(n-1)*[A(n-1)+1]/2}
=>
An={An*(An+1)/2}-{A(n-1)*[A(n-1)+1]/2}
=>
[A(n-1)]^2+A(n-1)=(An)^2-An
=>
[An-A(n-1)]*[An+A(n-1)]=An+A(n-1)
=>
An-A(n-1)=1
A1=1(自己算，不写了)
=>
An=n

Answer 2

n=1

Answer 3

bu dong