设数列{an}的前n项和Sn=2an-2n(1)证明数列{an+1-2an}是等差数列(2)证明数列{an+2}是等比数列
问题描述:
设数列{an}的前n项和Sn=2an-2n(1)证明数列{an+1-2an}是等差数列(2)证明数列{an+2}是等比数列
(3)求{an}的通项公式
答
Sn=2an-2n则Sn+1=2an+1-2(n+1)an+1=Sn+1-Sn=2an+1-2an-2则an+1-2an=2所以{an+1-2an}是等差数列(2)an+1-2an=2则an+1+2=2(an+2)所以{an+2}是等比数列(3)a1=S1=2a1-2a1=2因为{an+2}是公比为2的等比数列所以an+2...