问题
解答题
已知数列{an}的前n项和Sn=(n2+n)•3n. (Ⅰ)求
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答案
(1)lim n→∞
=an Sn lim n→∞
=Sn-Sn-1 Sn
(1-lim n→∞
)=1-Sn-1 Sn lim n→∞ Sn-1 Sn lim n→∞
=Sn-1 Sn lim n→∞
•n-1 n+1
=1 3
,所以1 3 lim n→∞
=an Sn
(6分)2 3
(2)当n=1时,
=S1=6>3;a1 12
当n>1时,
+a1 12
+…+a2 22
=an n2
+S1 12
+…+S2-S1 22 Sn-Sn-1 n2
=(
-1 12
) S1 +(1 22
-1 22
) S2 +…+(1 32
-1 (n-1)2
)Sn-1+1 n2
Sn>1 n2
Sn=1 n2
•3n>3nn2+n n2
所以,n≥1时,
+a1 12
+…+a2 22
>3n.(12分)an n2