已知数列{an}的前n项和Sn满足Sn=2an-1,等差数列{bn}满足b1=a1,b4=7.求数列{an}、{bn}的通项公式.

问题描述:

已知数列{an}的前n项和Sn满足Sn=2an-1,等差数列{bn}满足b1=a1,b4=7.求数列{an}、{bn}的通项公式.

∵Sn=2an-1,∴n≥2时,Sn-1=2an-1-1,∴两式相减可得,an=2an-2an-1,∴an=2an-1,n=1时,a1=2a1-1,∴a1=1,∴数列{an}是以1为首项,2为公比的等比数列,∴an=2n-1;∵b1=a1,b4=7,∴b1=1,公差为2,∴bn=1+(n-1...