题目内容
已知数列ξ中,满足a1=1且an+1=
,则
(n2an)=( )
| an |
| 1+nan |
| lim |
| n→∞ |
| A.1 | B.
| C.2 | D.
|
取倒数得:
-
=n
n分别取1,2,…n-1,累加得:
-
=1+2+…+n-1
∵a1=1
∴
=
∴an=
∴
(n2an)=
=2
故选C.
| 1 |
| an+1 |
| 1 |
| an |
n分别取1,2,…n-1,累加得:
| 1 |
| an |
| 1 |
| a1 |
∵a1=1
∴
| 1 |
| an |
| n2-n+2 |
| 2 |
∴an=
| 2 |
| n2-n+2 |
∴
| lim |
| n→∞ |
| lim |
| n→∞ |
| 2n2 |
| n2-n+2 |
故选C.
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