Question

若数列{an}的前n项和Sn,a1=2,且对任意大于1的整数n,点（根号下Sn,根号下Sn-1）在直线x-y-根号2＝0

Answer 1

根号下Sn-根号下S(n-1)-根号2＝0

根号下Sn-根号下S(n-1)=根号2

设bn=(Sn)^(1/2)

则：bn-b(n-1)=根号2

b1=(S1)^(1/2)=(a1)^(1/2)=根号2

bn=b1+(n-1)d=(根号2)n

Sn=2n^2

an=Sn-S(n-1)=2n^2-2(n-1)^2=2(2n+1)