a(n+1)=[f(√an)]^2,
a(n+1)=[√an/(√an+1)] ^2.
则√a(n+1)=√an/(√an+1)
1/√a(n+1)= (√an+1)/√an
即1/√a(n+1)=1+1/√an,
所以数列{1/√an }是等差数列,所以1/√an=1+(n-1)*1,
1/√an=n,an=1/n^2.