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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等差数列前n项和为Sn,若a(n+1)=3Sn,a1=1,则通项an?
![等差数列前n项和为Sn,若a(n+1)=3Sn,a1=1,则通项an?](/uploads/image/z/20203223-23-3.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%8B%A5a%EF%BC%88n%2B1%EF%BC%89%3D3Sn%2Ca1%3D1%2C%E5%88%99%E9%80%9A%E9%A1%B9an%3F)
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).
[[[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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