记数列{an}的前n项和为Sn,且Sn=2(an-1),则a2=_.

问题描述:

记数列{an}的前n项和为Sn,且Sn=2(an-1),则a2=______.

∵Sn=2(an-1),
∴S1=2(a1-1),
∴a1=2
∵S2=2(a2-1)=2+a2
∴a2=4
故答案为:4