已知数列{an}满足a1=
,an+1=
.
(1)证明数列
是等差数列,并求{an}的通项公式;
(2)若数列{bn}满足bn=
,求数列{bn}的前n项和Sn.
已知数列{an}满足a1=,an+1=.
(1)证明数列是等差数列,并求{an}的通项公式;
(2)若数列{bn}满足bn=,求数列{bn}的前n项和Sn.
(1)证明∵an+1=,
=2,
是等差数列,
+(n-1)×2=2+2n-2=2n,即an=
(2)解∵bn=
,
∴Sn=b1+b2+…+bn=1+
+…+
,
则Sn=
+…+
,
两式相减得Sn=1+
+…+
=2
,∴Sn=4-