已知正项数列{an}满足a1=1,Sn是数列{an}的前n项和,对任意n∈N+,有2Sn=2an^2+an-1.记bn=an/2^n

题目

已知正项数列{an}满足a1=1,Sn是数列{an}的前n项和,对任意n∈N+,有2Sn=2an^2+an-1.记bn=an/2^n
求数列bn的前n项和
先求an的通项

答案

2Sn-1=2an-1^2+an-1-1两式相减,得2Sn-2Sn-1=2an^2-2an-1^2+an-an-12an=2(an-an-1)(an+an-1)+an-an-10=2(an-an-1)(an+an-1)-(an+an-1)(an+an-1)(2an-2an-1-1)=0an=-an-1或2an=an-1 -1因为是正项,所以每项都是正数,第...