题目内容
(2008•温州模拟)等差数列{an}中,Sn是其前n项和,
-
=2,则
的值为( )
| S2008 |
| 2008 |
| S2006 |
| 2006 |
| lim |
| n→∞ |
| Sn |
| n2 |
分析:由等差数列的求和公式可得,
=a1+
,代入已知
-
=2可求公差d,再把所求的和代入
可求
| Sn |
| n |
| (n-1)d |
| 2 |
| S2008 |
| 2008 |
| S2006 |
| 2006 |
| lim |
| n→∞ |
| Sn |
| n2 |
解答:解:由等差数列的求和公式可得,Sn=na1+
∴
=a1+
∵
-
=a1+
-a1-
=d
∴d=2
∴
=
=
(
+1-
)=1
故选:B
| n(n-1)d |
| 2 |
∴
| Sn |
| n |
| (n-1)d |
| 2 |
∵
| S2008 |
| 2008 |
| S2006 |
| 2006 |
| 2007d |
| 2 |
| 2005d |
| 2 |
∴d=2
∴
| lim |
| n→∞ |
| Sn |
| n2 |
| lim |
| n→∞ |
| na1+n(n-1) |
| n2 |
| lim |
| n→∞ |
| a1 |
| n |
| 1 |
| n |
故选:B
点评:本题主要考查了数列极限的求解,此类极限的求解一般是在分式 的分子分母上同时除以一个因式,解题的关键是熟练应用等差数列的求和公式.
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