设数列{an}满足a1+a22+a322+…+an2n-1=2n,n∈N*.(1)求数列{an}的通项公式;(2)设bn=an(an-1)(an+1-1),求数列{bn}的前n项和Sn.

问题描述:

设数列{an}满足a1+

a2
2
+
a3
22
+…+
an
2n-1
=2n,n∈N*
(1)求数列{an}的通项公式;
(2)设bn=
an
(an-1)(an+1-1)
,求数列{bn}的前n项和Sn

(1)∵a1+a22+a322+…+an2n-1=2n,n∈N*,①∴当n=1时,a1=2.当n≥2时,a1+a22+a322+…+an-12n-2=2(n-1),②①-②得,an2n-1=2.∴an=2n.a1=2,适合上式,∴an=2n(n∈N*).(2)由(1)得an=2n.∴bn=an(an-1)(...