已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=n+12an+1(n∈N*).(Ⅰ)求数列{an}的通项公式;(Ⅱ)求数列{n2an}的前n项和Tn.-数学

题目简介

已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=n+12an+1(n∈N*).(Ⅰ)求数列{an}的通项公式;(Ⅱ)求数列{n2an}的前n项和Tn.-数学

题目详情

已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=
n+1
2
an+1(n∈N*)

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求数列{n2an}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵a1=1,a1+2a2+3a3+…+nan=class="stub"n+1
2
an+1(n∈N*)

a1+2a2+3a3+…+(n-1)an-1=class="stub"n
2
an

∴nan=class="stub"n+1
2
an+1-class="stub"n
2
an

an+1
an
=class="stub"3n
n+1

在a1=1,a1+2a2+3a3+…+nan=class="stub"n+1
2
an+1(n∈N*)

取n=1,得a2=1,
∴an+1=a2×
a3
a2
×
a4
a3
×…×
an+1
an

=1×(3×class="stub"2
3
)
×(3×class="stub"3
4
)
×…×(3×class="stub"n
n+1
)

=3n-1×class="stub"2
n+1

an=
1,n=1
3n-2•class="stub"2
n
,n≥2

(Ⅱ)∵an=
1,n=1
3n-2•class="stub"2
n
,n≥2

∴n2an=
1,n=1
2n•3n-2,n≥2

∴Tn=1+4×30+6×3+8×32+…+2n•3n-2,①
3Tn=3+4×3+6×32+8×33+…+2(n-1)•3n-2+2n•3n-1,②
①-②,得-2Tn=-2+4+2×(3+32+33+…+3n-2)-2n×3n-1
=2+2×
3(1-3n-2)
1-3
-2n×3n-1
=2+3n-1-3-2n×3n-1
=3n-1-1-2n×3n-1
∴Tn=class="stub"1
2
+n×3n-1-
3n-1
2

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