已知数列{an}的前n项和Sn=(n2+n)•3n.(Ⅰ)求limn→∞anSn;(Ⅱ)证明:a112+a222+…+ann2>3n.-数学

题目简介

已知数列{an}的前n项和Sn=(n2+n)•3n.(Ⅰ)求limn→∞anSn;(Ⅱ)证明:a112+a222+…+ann2>3n.-数学

题目详情

已知数列{an}的前n项和Sn=(n2+n)•3n
(Ⅰ)求
lim
n→∞
an
Sn
;(Ⅱ)证明:
a1
12
+
a2
22
+…+
an
n2
>3n
题型:解答题难度:中档来源:不详

答案

(1)
lim
n→∞
an
Sn
=
lim
n→∞
Sn-Sn-1
Sn
=
lim
n→∞
(1-
Sn-1
Sn
)=1-
lim
n→∞
Sn-1
Sn
lim
n→∞
Sn-1
Sn
=
lim
n→∞
class="stub"n-1
n+1
•class="stub"1
3
=class="stub"1
3
,所以
lim
n→∞
an
Sn
=class="stub"2
3
(6分)
(2)当n=1时,
a1
12
=S1=6>3

当n>1时,
a1
12
+
a2
22
+…+
an
n2
=
S1
12
+
S2-S1
22
+…+
Sn-Sn-1
n2

=(class="stub"1
12
-class="stub"1
22
S1 +(class="stub"1
22
-class="stub"1
32
S2 +…+(class="stub"1
(n-1)2
-class="stub"1
n2
)Sn-1+class="stub"1
n2
Sn>class="stub"1
n2
Sn
=
n2+n
n2
3n3n

所以,n≥1时,
a1
12
+
a2
22
+…+
an
n2
3n
.(12分)

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