设Sn是等比数列{an}的前n项和,且Sn=2an+n(1)证明:数列{an-1}是等比数列(2)数列{bn}满足bn=1/(2-an),证明:b1+b2+.+bn<1

问题描述:

设Sn是等比数列{an}的前n项和,且Sn=2an+n
(1)证明:数列{an-1}是等比数列
(2)数列{bn}满足bn=1/(2-an),证明:b1+b2+.+bn<1

(1)令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上,数列{an-1}是首项为a1-1=-2,公比为2的等比数列.(2)an-1=(a1-1)×qⁿ⁻¹=-...