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