已知在数列{an}中,a1=12,Sn是其前n项和,且Sn=n2an-n(n-1).(1)求{an}的通项公式;(2)令bn=(12)n+1-an,记数列{bn}的前n项和为Tn,求证:Tn<2.-数

题目简介

已知在数列{an}中,a1=12,Sn是其前n项和,且Sn=n2an-n(n-1).(1)求{an}的通项公式;(2)令bn=(12)n+1-an,记数列{bn}的前n项和为Tn,求证:Tn<2.-数

题目详情

已知在数列{an}中,a1=
1
2
,Sn是其前n项和,且Sn=n2an-n(n-1)
(1)求{an}的通项公式;
(2)令bn=(
1
2
)n+1-an
,记数列{bn}的前n项和为Tn,求证:Tn<2.
题型:解答题难度:中档来源:不详

答案

(1)∵an=Sn-Sn-1 (n≥2),Sn=n2an-n(n-1)
∴Sn=n2(Sn-Sn-1)-n(n-1),即(n2-1 )Sn-n2Sn-1=n(n-1),
class="stub"n+1
n
Sn
-class="stub"n
n-1
Sn-1
=1,∴{class="stub"n+1
n
Sn
}是首项为1,公差为1的等差数列
class="stub"n+1
n
Sn
=1+(n-1)×1=n,∴Sn=
n2
n+1

Sn=n2an-n(n-1)
n2
n+1
=n2an-n(n-1)

∴an=1-class="stub"1
n2+n

(2)证明:由(1)知,bn=(class="stub"1
2
)n+1-an
=(class="stub"1
2
)
n
+class="stub"1
n2+n
=(class="stub"1
2
)
n
+class="stub"1
n
-class="stub"1
n+1

∴Tn=class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
+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
2n
+1-class="stub"1
n+1
<2

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