设数列{an}的前 项和为Sn.已知a1=a,an+1=Sn+3n,n∈N*(Ⅰ)设bn=Sn-3n,求数列{bn}的通项公式;(Ⅱ)若an+1≥an,n∈N*,求a的取值范围.
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![设数列{an}的前 项和为Sn.已知a1=a,an+1=Sn+3n,n∈N*(Ⅰ)设bn=Sn-3n,求数列{bn}的通项公式;(Ⅱ)若an+1≥an,n∈N*,求a的取值范围.](/uploads/image/z/4532795-35-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8D+%E9%A1%B9%E5%92%8C%E4%B8%BASn%EF%BC%8E%E5%B7%B2%E7%9F%A5a1%EF%BC%9Da%2Can%2B1%EF%BC%9DSn%EF%BC%8B3n%2Cn%E2%88%88N%2A%28%E2%85%A0%29%E8%AE%BEbn%EF%BC%9DSn%EF%BC%8D3n%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%88%E2%85%A1%EF%BC%89%E8%8B%A5an%2B1%E2%89%A5an%2Cn%E2%88%88N%2A%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%EF%BC%8E)
设数列{an}的前 项和为Sn.已知a1=a,an+1=Sn+3n,n∈N*(Ⅰ)设bn=Sn-3n,求数列{bn}的通项公式;(Ⅱ)若an+1≥an,n∈N*,求a的取值范围.
设数列{an}的前 项和为Sn.已知a1=a,an+1=Sn+3n,n∈N*
(Ⅰ)设bn=Sn-3n,求数列{bn}的通项公式;(Ⅱ)若an+1≥an,n∈N*,求a的取值范围.
设数列{an}的前 项和为Sn.已知a1=a,an+1=Sn+3n,n∈N*(Ⅰ)设bn=Sn-3n,求数列{bn}的通项公式;(Ⅱ)若an+1≥an,n∈N*,求a的取值范围.
(1)an+1=Sn+3n
当n大于2时 sn-sn-1=an
an=Sn-1+3(n-1)
两等式相减an+1-an=sn-sn-1+3
an+1+3=2(an+3)
数列{an+3}是以a+3为首项2为公比的等比数列
an+3=(a+3)*2^n-1
Bn=Sn-3n=an+1-6n=(a+3)*2^n-6n-3
(2)an+1≥an an=(a+3)*2^n-1-3
an+1=(a+3)*2^n-3
所以2(a+3)>a+3
a>-3
:(Ⅰ)依题意,Sn 1-Sn=an 1=Sn 3n,即Sn 1=2Sn 3n,
由此得Sn 1-3n 1=2Sn 3n-3n 1=2(Sn-3n).(4分)
因此,所求通项公式为bn=Sn-3n=(a-3)2n-1,n∈N*.①(6分)
(Ⅱ)由①知Sn=3n (a-3)2n-1,n∈N*,
于是,当n≥2时,
an=Sn-Sn-1=3n (a-3)×2n...
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:(Ⅰ)依题意,Sn 1-Sn=an 1=Sn 3n,即Sn 1=2Sn 3n,
由此得Sn 1-3n 1=2Sn 3n-3n 1=2(Sn-3n).(4分)
因此,所求通项公式为bn=Sn-3n=(a-3)2n-1,n∈N*.①(6分)
(Ⅱ)由①知Sn=3n (a-3)2n-1,n∈N*,
于是,当n≥2时,
an=Sn-Sn-1=3n (a-3)×2n-1-3n-1-(a-3)×2n-2=2×3n-1 (a-3)2n-2,
an 1-an=4×3n-1 (a-3)2n-2=2n−2[12•(
3
2
)n−2 a−3],
当n≥2时,an 1≥an⇔12•(
3
2
)n−2 a−3≥0⇔a≥-9.
又a2=a1 3>a1.
综上,所求的a的取值范围是[-9, ∞).
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