等差数列前n项和为Sn,若a（n+1）=3Sn,a1=1,则通项an?

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递推式:a(n+1)=3Sn, a1=1
[[[1]]
先计算前4项.
a2=3S1=3a1=3
a3=3S2=12
a4=3S3=48
∴a1=1, a2=3 a3=12, a4=48
[[[2]]]
a(n+1)=3Sn
an=3S(n-1). n≥2
两式相减,可得
[a(n+1)]-(an)=3an
∴a(n+1)=4an
∴结合上面可知
当n≥2时,数列{an}是公比=4的等比数列
通项:
n=1时, a1=1
n≥2时, an=3×4^(n-2).

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