14.对任意x属于R,函数f(x)满足f(x+1)={√f(x)-[f(x)]^2}+1/2,设an=[f(n)]^2-f(n),数列{an}的前15贡和为期不远31/16,则f(15)=.14.对任意x属于R，函数f(x)满足f(x+1)={√f(x)-[f(x)]^2}+1/2,设an=[f(n)]^2-f(n),数列{an}

Question

14.对任意x属于R,函数f(x)满足f(x+1)={√f(x)-[f(x)]^2}+1/2,设an=[f(n)]^2-f(n),数列{an}的前15贡和为期不远3 1/16,则f(15)=__________.
14.对任意x属于R，函数f(x)满足f(x+1)={√f(x)-[f(x)]^2}+1/2,设an=[f(n)]^2-f(n),数列{an}的前15和为-31/16，则f(15)=__________.

Answer 1

由f(n+1)={f(n)-[f(n)]2}^0.5+1/2=(-an)^0.5+0.5和a(n+1)=[f(n+1)]^2-f(n+1),
可得：a(n+1)=-an-1/4,即a(n+1)+an=-1/4；
S15=a15+(a14+a13)+……+(a2+a1)=a15+7*(-1/4)=-31/16
=a15+7*(-1/4)=-31/16,解得a15=-3/16,
又因a(n+1)+an=-1/4,得a14=-1/16,即[f(14)]^2-f(14)=-1/16,带入f(x+1)={√f(x)-[f(x)]^2}+1/2,得到f(15)=√[-a14]+1/2=3/4