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+
1
2
}为等比数列;
(2)记Tn=S1+S2+L+Sn,求Tn的表达式;
(3)记Cn=
2
3
(an+
1
2
),求数列{nCn}的前n项和Pn
(1)∵3an=2Sn+n,
∴a1=1,
当n≥2时,3(an-an-1)=2an+1,即an=3an-1+1,
∴an+
1
2
=3an-1+1+
1
2
=3(an-1+
1
2
),
∴数列{an+
1
2
}是首项为
3
2
,公比为3的为等比数列;
(2)由(1)知,an+
1
2
=
3
2
•3n-1
∴an=
1
2
×3n-
1
2

∴Sn=a1+a2+…+an
=
1
2
3(1-3n)
1-3
-
n
2

=
3
4
•3n-
1
4
(2n+3),
∴Tn=S1+S2+…+Sn
=
3
4
(3+32+…+3n)-
1
4
×
(5+2n+3)n
2

=
3
4
3(1-3n)
1-3
-
n(n+4)
4

=
9
8
(3n-1)-
n(n+4)
4

(3)∵Cn=
2
3
(an+
1
2
)=
2
3
×
1
2
×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
=
1-3n
1-3
-n•3n
=
1-2n
2
×3n-
1
2

∴Pn=
1+(2n-1)•3n
4
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