已知数列{an}满足a1=3,且an+1-3an=3n,(n∈N*),数列{bn}满足bn=3-nan.
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已知数列{an}满足a1=3,且an+1-3an=3n,(n∈N*),数列{bn}满足bn=3-nan.
(1)求证:数列{bn}是等差数列;
(2)设S
(1)求证:数列{bn}是等差数列;
(2)设S
(1)证明:由bn=3-nan得an=3nbn,则an+1=3n+1bn+1.
代入an+1-3an=3n中,得3n+1bn+1-3n+1bn=3n,
即得bn+1-bn=
1
3.
所以数列{bn}是等差数列.(6分)
(2) 因为数列{bn}是首项为b1=3-1a1=1,公差为
1
3等差数列,
则bn=1+
1
3(n-1)=
n+2
3,则an=3nbn=(n+2)×3n-1.(8分)
从而有
an
n+2=3n-1,
故Sn=
a1
3+
a2
4+
a3
5++
an
n+2=1+3+32++3n-1=
1-3n
1-3=
3n-1
2.(11分)
则
Sn
S2n=
3n-1
32n-1=
1
3n+1,
由
1
128<
Sn
S2n<
1
4,得
1
128<
1
3n+1<
1
4.
即3<3n<127,得1<n≤4.
故满足不等式
1
128<
Sn
S2n<
1
4的所有正整数n的值为2,3,4.(14分)
代入an+1-3an=3n中,得3n+1bn+1-3n+1bn=3n,
即得bn+1-bn=
1
3.
所以数列{bn}是等差数列.(6分)
(2) 因为数列{bn}是首项为b1=3-1a1=1,公差为
1
3等差数列,
则bn=1+
1
3(n-1)=
n+2
3,则an=3nbn=(n+2)×3n-1.(8分)
从而有
an
n+2=3n-1,
故Sn=
a1
3+
a2
4+
a3
5++
an
n+2=1+3+32++3n-1=
1-3n
1-3=
3n-1
2.(11分)
则
Sn
S2n=
3n-1
32n-1=
1
3n+1,
由
1
128<
Sn
S2n<
1
4,得
1
128<
1
3n+1<
1
4.
即3<3n<127,得1<n≤4.
故满足不等式
1
128<
Sn
S2n<
1
4的所有正整数n的值为2,3,4.(14分)
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