已知数列{an}满足a1=3,an+1−3an=3n(n∈N*),数列{bn}满足bn=an3n.
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已知数列{an}满足a1=3,a
解(1)证明:由bn=
an
3n,得bn+1=
an+1
3n+1,
∴bn+1−bn=
an+1
3n+1−
an
3n=
1
3---------------------(2分)
所以数列{bn}是等差数列,首项b1=1,公差为
1
3-----------(4分)
∴bn=1+
1
3(n−1)=
n+2
3------------------------(6分)
(2)an=3nbn=(n+2)×3n−1-------------------------(7分)
∴Sn=a1+a2+…+an=3×1+4×3+…+(n+2)×3n-1----①
∴3Sn=3×3+4×32+…+(n+2)×3n-------------------②(9分)
①-②得−2Sn=3×1+3+32+…+3n−1−(n+2)×3n
=2+1+3+32+…+3n-1-(n+2)×3n=
3n+3
2−(n+2)×3n------(11分)
∴Sn=−
3n+3
4+
(n+2)3n
2-----------------(12分)
an
3n,得bn+1=
an+1
3n+1,
∴bn+1−bn=
an+1
3n+1−
an
3n=
1
3---------------------(2分)
所以数列{bn}是等差数列,首项b1=1,公差为
1
3-----------(4分)
∴bn=1+
1
3(n−1)=
n+2
3------------------------(6分)
(2)an=3nbn=(n+2)×3n−1-------------------------(7分)
∴Sn=a1+a2+…+an=3×1+4×3+…+(n+2)×3n-1----①
∴3Sn=3×3+4×32+…+(n+2)×3n-------------------②(9分)
①-②得−2Sn=3×1+3+32+…+3n−1−(n+2)×3n
=2+1+3+32+…+3n-1-(n+2)×3n=
3n+3
2−(n+2)×3n------(11分)
∴Sn=−
3n+3
4+
(n+2)3n
2-----------------(12分)
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