已知数列{an}的前n项和为Sn,且点在直线2x-y-3=0上,则数列{an}的通项公式为________.
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3?2n-1
解析分析:将点(an,Sn)的坐标代入直线方程2x-y-3=0可得2an-Sn=3,下推一项,两者作差,整理即可.
解答:∵点(an,Sn)在直线2x-y-3=0上,∴2an-Sn=3,①∴2an-1-Sn-1=3(n≥2)②①-②得:2(an-an-1)=Sn-Sn-1=an,∴an=2an-1(n≥2)又2a1-a1=3,∴a1=3.∴数列{an}为a1=3,公比q=2的等比数列,∴an=3?2n-1.故