知数列{an}的前n项和为Sn.满足an+Sn=2n 求an;设bn=(2-n)(an-2),bn属于(-无穷,m),求m取值范围

问题描述:

知数列{an}的前n项和为Sn.满足an+Sn=2n 求an;设bn=(2-n)(an-2),bn属于(-无穷,m),求m取值范围

an+Sn=2n ,an-1+Sn-1=2n-2两式相减,2an-an-1=2用待定系数法2an+p=an-1+2+p2(an+p/2)/(an-1+2+p)=1令p/2=2+p,p=-4,那么(an-2)/(an-1-2)=1/2那么数列(an-2)是以a1-2为首项1/2为公比的等比数列,而a1可以算出来是1,即an-...