已知数列{an}的前n项和为Sn,满足an+Sn=2n. (Ⅰ)证明:数列{an-2}为等比数列,并求出an; (Ⅱ)设bn=(2-n)(an-2),求{bn}的最大项.

问题描述:

已知数列{an}的前n项和为Sn,满足an+Sn=2n.
(Ⅰ)证明:数列{an-2}为等比数列,并求出an
(Ⅱ)设bn=(2-n)(an-2),求{bn}的最大项.

(Ⅰ)证明:由a1+s1=2a1=2得a1=1;由an+Sn=2n得an+1+Sn+1=2(n+1)两式相减得2an+1-an=2,即2an+1-4=an-2,即an+1-2=12(an-2)是首项为a1-2=-1,公比为12的等比数列.故an-2=-(12)n−1,故an=2-(12)n−1,.(Ⅱ)...