问题 解答题
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
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→∞
n-1
n+1
1
3
=
1
3
,所以
lim
n→∞
an
Sn
=
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

=(

1
12
-
1
22
S1 +(
1
22
-
1
32
S2 +…+(
1
(n-1)2
-
1
n2
)Sn-1+
1
n2
Sn
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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