数列{an}中,a1=1,Sn+1=4an+2设bn=an+1-2an,求证{bn}是等比数列,并求其通项.

问题描述:

数列{an}中,a1=1,Sn+1=4an+2设bn=an+1-2an,求证{bn}是等比数列,并求其通项.

∵S(n+1)=4an+2∴当n≥2时,Sn=4a(n-1)+2∴S(n+1)-Sn=4an-4a(n-1),即:a(n+1)=4an-4a(n-1).(1)∴a(n+1)-2an=2[an-2a(n-1)],即:bn=2b(n-1).∴{bn}是等比数列.等比数列{bn}的公比是2.首项b1=a2-2a1,又S2=4a1+2,a1+a2...