名校
解题方法
1 . 设
条直线最多把平面分成
部分,其求法如下:易知一条直线最多把平面分成
部分,两条直线最多把平面分成
部分,3条直线分平面,要使所得部分尽量多,则第三条直线必与前两条直线都相交,产生2个交点,这2个交点都在第3条直线上,并把第三条直线分成3段,这3段的每一段都在
部分的某部分中,它把所在部分一分为二,故增加了3部分,即
,依次类推得
,累加化简得
.根据上面的想法,设
个平面最多把空间分成
部分,且![](https://staticzujuan.xkw.com/quesimg/Upload/formula/76a16dcc0f10910231b301311a3f4f02.png)
(1)求出![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a548938d87c80ac47910607d3857007f.png)
(2)写出
与
之间的递推关系式
(3)求出数列
的通项公式
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/96abfe2da27a63e6affb19a0c80236d9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/039e4fe671d61e59b96ee525c9df43e8.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/18d8e8f821111de8075e5c3dfb22a5d6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3e88093a749c0d46e0ee931ecfaff925.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b8c82f3e67cbea29ba76cf23b2fd7dee.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2eed59e4b6dceac18723abab5fadcf6d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f512fe4f79cdf9f8f21d7c7075359b70.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/686ece75006ad358f23314dc8a246e11.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/76a16dcc0f10910231b301311a3f4f02.png)
(1)求出
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a548938d87c80ac47910607d3857007f.png)
(2)写出
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e95931effbd59c43e8ed1ea09962b84f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/686ece75006ad358f23314dc8a246e11.png)
(3)求出数列
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5fce83115a50f99e08e9a2db7267aeed.png)
您最近一年使用:0次