已知{an}是正数组成的数列,a1=1,且点()(nN*)在函数y=x2+1的图象上.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)若数列{bn}满足b1=1,bn+1=bn+,求证:bn?bn+2<b2n+1.