已知数列{an}的前n项和Sn满足2Sn2=2anSn-an(n≥2)且a1=2,求an和Sn.

问题描述:

已知数列{an}的前n项和Sn满足2Sn2=2anSn-an(n≥2)且a1=2,求an和Sn

∵2Sn2=2anSn-an(n≥2),∴2Sn2=2(Sn-Sn-1)Sn-(Sn-Sn-1).计算化简得,Sn-1-Sn=2SnSn-1,两边同除以SnSn-1,得1Sn−1Sn−1=2,(n≥2),∴数列{1Sn}是以2为公差的等差数列,首项1S1=1a1=12,∴数列{1Sn}的通...