已知:数列{a
n}的前n项和S
n=n
2+2n(n∈N
*)
(1)求:通项
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824044623752348.png)
(2)求和:
(1) a
n=" 2n+1;(2)"
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824044623862698.png)
.
试题分析:(1)利用
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240446238771478.png)
,即可求出结果;
(2)由于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240446238931788.png)
,所以求
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240446238461024.png)
可以利用裂项相消法求和即可 .
试题解析:解:(Ⅰ)当n≥2时,a
n=S
n-S
n-1=2n+1, 2分
n=1时,a
1=S
1=3适合上式 3分
∴a
n=2n+1, n∈N
*, 4分
(Ⅱ)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240446238931788.png)
6分
∴原式
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240446239241395.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824044623940927.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824044623862698.png)
8分
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