已知函数,f(x)=
,数列{an}满足a1=1,an+1=f(an)(n∈N*)
(I)求证数列{
}是等差数列,并求数列{an}的通项公式;
(II)记Sn=a1a2+a2a3+..anan+1,求Sn.
0 91969 91977 91983 91987 91993 91995 91999 92005 92007 92013 92019 92023 92025 92029 92035 92037 92043 92047 92049 92053 92055 92059 92061 92063 92064 92065 92067 92068 92069 92071 92073 92077 92079 92083 92085 92089 92095 92097 92103 92107 92109 92113 92119 92125 92127 92133 92137 92139 92145 92149 92155 92163 266669
(I)求证数列{
(II)记Sn=a1a2+a2a3+..anan+1,求Sn.