题目内容
数列{an}的前n项和为Sn,a1=2,Sn=
an+1-1(n∈N*).
(Ⅰ)求a2,a3;
(Ⅱ)求数列{an}的通项an;
(Ⅲ)求数列{nan}的前n项和Tn.
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(Ⅰ)求a2,a3;
(Ⅱ)求数列{an}的通项an;
(Ⅲ)求数列{nan}的前n项和Tn.
(Ⅰ)∵a1=2,Sn=
an+1-1(n∈N*),
∴当n=1时,S1=
a2-1=a1=2,
解得a2=6.
当n=2时,S2=
a3-1=2+6=8,
解得a3=18.
(Ⅱ)∵a1=2,Sn=
an+1-1(n∈N*),
∴当n≥2时,Sn=
an+1-1,Sn-1=
an-1,
∴an=Sn-Sn-1=
an+1-
an,
即an+1=3an.
对于a2=3a1也满足上式,
∴数列{an}是首项为2,公比为3的等比数列,
∴an=2•3n-1(n∈N*).
( III)∵an=2•3n-1(n∈N*),
∴nan=2n•3n-1,
∴Tn=2•1+4•3+6•32+8•33+…+2n•3n-1,
3Tn=2•3+4•32+6•33+8•34+…+2n•3n,
相减得,-2Tn=2(1+3+32+33+…+3n-1)-2n•3n
=2•
-2n•3n
=3n-1-2n•3n,
∴Tn=
.
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∴当n=1时,S1=
1 |
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解得a2=6.
当n=2时,S2=
1 |
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解得a3=18.
(Ⅱ)∵a1=2,Sn=
1 |
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∴当n≥2时,Sn=
1 |
2 |
1 |
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∴an=Sn-Sn-1=
1 |
2 |
1 |
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即an+1=3an.
对于a2=3a1也满足上式,
∴数列{an}是首项为2,公比为3的等比数列,
∴an=2•3n-1(n∈N*).
( III)∵an=2•3n-1(n∈N*),
∴nan=2n•3n-1,
∴Tn=2•1+4•3+6•32+8•33+…+2n•3n-1,
3Tn=2•3+4•32+6•33+8•34+…+2n•3n,
相减得,-2Tn=2(1+3+32+33+…+3n-1)-2n•3n
=2•
1-3n |
1-3 |
=3n-1-2n•3n,
∴Tn=
(2n-1)•3n+1 |
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