Question

已知数列an的前n项和为Sn,a1=1,Sn=2an+1,则Sn=?

Answer 1

Sn-1 = 2an-1 + 1
an = Sn - Sn-1 = 2an - 2an-1，
an = 2an-1 = …… =2^n-1，
Sn = 2^n -1。
题目有一定问题，应为Sn = 2an - 1或原式只对n>=2成立。

Answer 2

∵ Sn=2an+1
∴S(n+1)=2a(n+1)+1
∴a(n+1)=S(n+1)-Sn
=2a(n+1)+1-(2an+1)
=2a(n+1)-2an
∴a(n+1)=2an
∴a(n+1)/an=2
∴{an}是等比数列，公比为2，首项为1
∴an=2^(n-1)
∴Sn=2^n-1

Answer 3

解
∵a(n+1)=S(n+1)-Sn
Sn=2a(n+1)=2[S(n+1)-Sn]
3Sn=2S(n+1)
S(n+1)/Sn=3/2
S1=1
∴Sn=S1*(3/2)^(n-1)=(3/2)^(n-1) (n>=1)