已知数列的前n项和为Sn,且a1=1,S(n+1)=4an+2,(1)设bn=a(n+1)--2n,求证bn是等比数列,

问题描述:

已知数列的前n项和为Sn,且a1=1,S(n+1)=4an+2,(1)设bn=a(n+1)--2n,求证bn是等比数列,
2)求Sn=a1+a2+a3.+an

证明:(Ⅰ)由于Sn+1=4an+1,①当n≥2时,Sn=4an-1+1.②①-②得an+1=4an-4an-1.所an+1-2an=2(an-2an-1).又bn=an+1-2an,所以bn=2bn-1.因为a1=1,且a1+a2=4a1+1,所以a2=3a1+1=4.所以b1=a2-2a1=2.故数列{bn}是首...