An=f【A(n-1)】=根号【A(n-1)²+1】,即An²=A(n-1)²+1,可得出An=根号nbn=an²,则bn=nSn=n(1+n)拆轮/2有不明白的地方再追碧手问,望采旅慧信纳
f(x)=根号(x^2+1)an=根号(a(n-1)^2+1)an^2=a(n-1)^2+1又bn=an^2所以b(n-1)=a(n-1)^2所以局盯敏bn=b(n-1)+1且b1=a1^2=1所以bn是以1为首项1公差桐枝的等则老差数列有bn=nsn=1/2n(n+1)=1/2n^2+1/2n