已知数列{an}的首项为1,前n项和为Sn,且满足an+1=3Sn,n∈N*.数列{bn}满足bn=log4an.
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已知数列{an}的首项为1,前n项和为Sn,且满足an+1=3Sn,n∈N*.数列{bn}满足bn=log4an.
(I)求数列{an}的通项公式;
(II)当n≥2时,试比较b1+b2+…+bn与
(n−1)
(I)求数列{an}的通项公式;
(II)当n≥2时,试比较b1+b2+…+bn与
1 |
2 |
![已知数列{an}的首项为1,前n项和为Sn,且满足an+1=3Sn,n∈N*.数列{bn}满足bn=log4an.](/uploads/image/z/3128834-2-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E9%A1%B9%E4%B8%BA1%EF%BC%8C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%EF%BC%8C%E4%B8%94%E6%BB%A1%E8%B6%B3an%2B1%3D3Sn%EF%BC%8Cn%E2%88%88N%2A%EF%BC%8E%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%3Dlog4an%EF%BC%8E)
(I)由an+1=3Sn(1),得an+2=3Sn+1(2),
由(2)-(1)得an+1-an+1=3an+1,
整理,得
an+2
an+1=4,n∈N*.
所以,数列a2,a3,a4,…,an,是以4为公比的等比数列.
其中,a2=3S1=3a1=3,
所以an=
1n=1
3•4n−2n≥2,n∈N*;
(II)由题意,bn=
0n=1
log43+(n−2)n≥2,n∈N*.
当n≥2时,b1+b2+b3+…+bn
=0+(log43+0)+(log43+1)+…+(log43+n-2)
=(n−1)log43+
1
2(n−2)(n−1)
=
n−1
2[2log43−1+(n−1)]
=
n−1
2[log4
9
4+(n−1)]>
(n−1)2
2,
所以b1+b2+b3++bn>
(n−1)2
2.
由(2)-(1)得an+1-an+1=3an+1,
整理,得
an+2
an+1=4,n∈N*.
所以,数列a2,a3,a4,…,an,是以4为公比的等比数列.
其中,a2=3S1=3a1=3,
所以an=
1n=1
3•4n−2n≥2,n∈N*;
(II)由题意,bn=
0n=1
log43+(n−2)n≥2,n∈N*.
当n≥2时,b1+b2+b3+…+bn
=0+(log43+0)+(log43+1)+…+(log43+n-2)
=(n−1)log43+
1
2(n−2)(n−1)
=
n−1
2[2log43−1+(n−1)]
=
n−1
2[log4
9
4+(n−1)]>
(n−1)2
2,
所以b1+b2+b3++bn>
(n−1)2
2.
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