题目内容
已知数列{an},Sn是其n前项的和,且满足3an=2Sn+n(n∈N*)
(1)求证:数列{an+
}为等比数列;
(2)记Tn=S1+S2+L+Sn,求Tn的表达式;
(3)记Cn=
(an+
),求数列{nCn}的前n项和Pn.
(1)求证:数列{an+
1 |
2 |
(2)记Tn=S1+S2+L+Sn,求Tn的表达式;
(3)记Cn=
2 |
3 |
1 |
2 |
(1)∵3an=2Sn+n,
∴a1=1,
当n≥2时,3(an-an-1)=2an+1,即an=3an-1+1,
∴an+
=3an-1+1+
=3(an-1+
),
∴数列{an+
}是首项为
,公比为3的为等比数列;
(2)由(1)知,an+
=
•3n-1,
∴an=
×3n-
,
∴Sn=a1+a2+…+an
=
•
-
=
•3n-
(2n+3),
∴Tn=S1+S2+…+Sn
=
(3+32+…+3n)-
×
=
•
-
=
(3n-1)-
.
(3)∵Cn=
(an+
)=
×
×3n=3n-1,
∴Pn=1×30+2×3+3×32+…+n•3n-1,
∴3Pn=1×3+2×32+…+(n-1)•3n-1+n•3n,
两式相减得:
-2Pn=1+3+32+…+3n-1-n•3n
=
-n•3n
=
×3n-
,
∴Pn=
.
∴a1=1,
当n≥2时,3(an-an-1)=2an+1,即an=3an-1+1,
∴an+
1 |
2 |
1 |
2 |
1 |
2 |
∴数列{an+
1 |
2 |
3 |
2 |
(2)由(1)知,an+
1 |
2 |
3 |
2 |
∴an=
1 |
2 |
1 |
2 |
∴Sn=a1+a2+…+an
=
1 |
2 |
3(1-3n) |
1-3 |
n |
2 |
=
3 |
4 |
1 |
4 |
∴Tn=S1+S2+…+Sn
=
3 |
4 |
1 |
4 |
(5+2n+3)n |
2 |
=
3 |
4 |
3(1-3n) |
1-3 |
n(n+4) |
4 |
=
9 |
8 |
n(n+4) |
4 |
(3)∵Cn=
2 |
3 |
1 |
2 |
2 |
3 |
1 |
2 |
∴Pn=1×30+2×3+3×32+…+n•3n-1,
∴3Pn=1×3+2×32+…+(n-1)•3n-1+n•3n,
两式相减得:
-2Pn=1+3+32+…+3n-1-n•3n
=
1-3n |
1-3 |
=
1-2n |
2 |
1 |
2 |
∴Pn=
1+(2n-1)•3n |
4 |
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