设数列n2an的前n项和为Sn,且Sn=n(n+1)(n+2),n∈N*.(1)求数列an的通项公式;(2)若数列bn满足bn=a1a2a3…an,n∈N*,求数列bn的通项公式及前n项和Tn;(3)

题目简介

设数列n2an的前n项和为Sn,且Sn=n(n+1)(n+2),n∈N*.(1)求数列an的通项公式;(2)若数列bn满足bn=a1a2a3…an,n∈N*,求数列bn的通项公式及前n项和Tn;(3)

题目详情

设数列n2an的前n项和为Sn,且Sn=n(n+1)(n+2),n∈N*
(1)求数列an的通项公式;
(2)若数列bn满足bn=a1a2a3…an,n∈N*,求数列bn的通项公式及前n项和Tn
(3)在(2)的条件下,求证:
3
b1
+
32
2b2
+
33
3b3
+…+
3n
nbn
=
n
n+1
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,a1=6;
当n≥2时,Sn=n(n+1)(n+2)①
Sn-1=(n-1)n(n+1)②
由①-②得:n2an=3n(n+1),即an=
3(n+1)
n

综上得:an=
3(n+1)
n
.(4分)
(2)因为an=
3(n+1)
n

所以bn=a1a2a3an=class="stub"3×2
1
×class="stub"3×3
2
×class="stub"3×4
3
××
3(n+1)
n
=3n(n+1)

故bn=3n(n+1).(6分)
Tn=2•3+3•32+4•33+…+n•3n-1+(n+1)•3n.③
3Tn=2•32+3•33+4•34+…+n•3n+(n+1)•3n+1.④
③-④得:-2Tn=2•3+32+33+…+3n-(n+1)•3n+1=class="stub"1
2
3n+1+ class="stub"3
2
-(n+1)•3n+1

化简得:Tn=(class="stub"n
2
+class="stub"1
4
)•3n+1-class="stub"3
4
.(9分)
(3)由bn=3n(n+1),得
3n
bn
=class="stub"1
n+1
,等式两端同时乘以class="stub"1
n

3n
nbn
=class="stub"1
n(n+1)
.则有
class="stub"3
b1
+
32
2b2
33
3b3
+…+
3n
nbn
=class="stub"1
1×2
+class="stub"1
2×3
+class="stub"1
3×4
+…class="stub"1
n(n+1)


1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+ class="stub"1
n
-class="stub"1
n+1

=1-class="stub"1
n+1

=class="stub"n
n+1
(12分)

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