对任意x属于R,函数f(x)满足f(x+1)=根号下{2f(x)-[f(x)]2}+1,设an=[f(n)]2-2f(n

f(x+1)=√[2f(x)-f(x)²] +1
f(x+1)-1=√[2f(x)-f(x)²]
算术平方根有意义,f(x+1)-1≥0 f(x+1)≥1
x为任意实数,x+1为任意实数,f(x)≥1
2f(x)-f(x)²≥0 f(x)[2-f(x)]≥0
f(x)≥1,2-f(x)≥0 f(x)≤2
综上,得1≤f(x)≤2
[f(x+1)-1]²=2f(x)-f(x)²
f(x+1)²-2f(x+1)+1=2f(x)-f(x)²
[f(x+1)²-2f(x+1)]+[f(x)²-2f(x)]=-1,为定值.
an=f(n)²-2f(n),数列{an}是等和数列.(不是等差数列)
数列奇数项=a1,偶数项=a2
S2013=a1+a2+a3+a4+...+a2013
=(a1+a2)+(a3+a4)+...+(a2011+a2012)+a2013
=(-1)×2012/2 +a2013
=-1006+a2013=-4027/4
a2013=1006-4027/4=-3/4
f(2013)²-2f(2013)=-3/4
整理,得
4f(2013)²-8f(2013)+3=0
[2f(2013)-1][2f(2013)-3]=0
f(2013)=1/2(1≤f(x)≤2,舍去)或f(2013)=3/2
f(2013)=3/2